Is there an orientable $3$-manifold with non-vanishing $w_2$? 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?
 A: 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.
A: 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.
