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In the case that $M$ is a closed orientable $3$-manifold, using Wu's formula we can show $w_1(M) =0 \implies w_2(M) =0$, and so $w_3 = w_1w_2 + Sq^1 w_2 = 0$ (or you can use the fact that $\chi(M)=0$ for closed orientable manifolds with odd dimension). It can then be shown that in fact $M$ is parallelizable and orientedly null-bordant.

If $M$ is compact and orientable, then its boundary is an orientable surface and therefore bounds so we can complete $M$ to a closed manifold $\bar{M}$ where the same argument applies, and we can again compute $w_2(M) = 0$.

Therefore if we want an example of an orientable $3$-manifold with $w_2(M)\neq 0$ it needs to be non-compact. Does anyone know an example?

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According to R. Kirby in The Topology of 4-Manifold, section VIII, Theorem 1 on page 46

Every orientable 3-manifold $M^3$ is spin and hence parallelizable.

First he proves the case when $M$ is compact. Then to complete the argument, in the case that $M$ is non-compact, he writes

If $T_M$ is non-trivial, then it is non-trivial on some compact piece of $M^3$, contracting the above.

Where "the above argument" is what I include as the "below argument":

Assume $M$ is compact and closed, possibly by moving to its double. Assume $w_2(M)\neq 0\in H^2(M;\mathbb{Z}_2)$ and let $C$ be a circle in $M$ which is Poincare dual to this class. Then $M\setminus C$ has a spin structure $\sigma$ which does not extend to $M$ and there is a dual surface $F^2\subseteq M$ which intersects $C$ transversely in one point $p$ ($F$ may be non-orientable). Then the total space of the normal disc bundle $\nu(F)$ to $F$ is isomorphic to the total space of the normal disc bundle to an immersion of $F$ in $\mathbb{R}^3$ and so has a spin structure. However spin structures on $\nu(F)$ are classified by $H^1(F;\mathbb{Z}_2)\cong H^1(F\setminus p;\mathbb{Z}_2)$, and this latter group classifies the spin structures on the normal disc bundle $\nu(F\setminus p)$. Thus $\sigma$ gives a spin structure on $\nu(F\setminus p)$ which must agree by restriction with that on $\nu(F)$. It follows from this that $\sigma$ must extend across $C$, which contradicts the assumption that $w_2(M)\neq 0$. Hence $M$ must be spin. And since $\pi_2SO_3=0$, it must be parallelizable.

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    $\begingroup$ Do you want to explain how to justify the italicized sentence? It is not hard and works for arbitrary bundles. $\endgroup$ – Moishe Kohan Apr 4 '19 at 0:01
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    $\begingroup$ Hi @MoisheKohan, every smooth manifold is CW and so can be written (in $Top$) as a colimit over its (compact) cells. Since the problem is topological it suffices to answer it in this category. I'll note that this argument does not work for "arbitrary bundles" due to the possible presence of phantom maps. Since the tangent bundle can always be classified by a map into a finite-dimensional Grassmannian these particular difficulties do not enter into the current picture. $\endgroup$ – Tyrone Apr 4 '19 at 9:22
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    $\begingroup$ @MoisheKohan, here mathoverflow.net/questions/139668/… , Anders And constructs a smooth 3-manifold admitting a complex line bundle classified by a phantom map, $\endgroup$ – Tyrone Apr 4 '19 at 15:53
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    $\begingroup$ @MichaelAlbanese, the statement is a direct quote from the reference. He includes nothing more, so Kirby's full intent is maybe a little unclear. As I understand it one should choose an exhaustion $M=\bigcup M_i$ of $M$ by compact submanifolds (see for example math.stackexchange.com/questions/1360900/…). Then each $M_i\subseteq M_{i+1}$ is a cofibration (choose a tubular neighbourhood) and $M$ is the colimit and homotopy colimit of the sequence. Then a map $M\rightarrow BSO_3$ classifying the tangent bundle is the colimit of the $f|_{M_i}$. $\endgroup$ – Tyrone Apr 6 '19 at 17:03
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    $\begingroup$ So when mapping out of a mapping telescope we get an exact sequence involving $\lim$ and $\lim^1$ terms. Essentially the phantom maps are the $\lim^1$ terms. As should be clear, they can only appear when when $X$ is an infinite complex, and in the case at hand, this can only happen if $\pi_1X$ is suitably disagreeable. There is a nice- and relevant - example of a phantom in the link I left above. $\endgroup$ – Tyrone Apr 7 '19 at 13:14
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All orientable three-manifolds $M$ are parallelizable. If you just want to deduce the noncompact case from the closed one, this requires little machinery.

Basically, you find first an exhaustion of $M$ with connected compact manifolds with boundary $M_k$. Then you inductively construct linearly independent vector fields $X,Y,Z$ on each $M_k$. Extending them from $M_k$ to $M_{k+1}$ is not trivial, and can be done e.g. taking an appropriate harmonic extension. See my answer to a very similar question here.

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