Integral class mod 2 reduction, and Stiefel Whitney class If there is an integral 3rd cohomology class called Y, such that whose mod 2 reduction is the Steenrod square of the 2nd cohomology of Stiefel Whitney class, then we have
$$ Y=Sq^1 w_2 \mod 2$$


*

*In 5 dim, is this true that such a lift from $Sq^1 w_2$ to the integral class $Y$
$$
\int_{5d} w_2Y=0?
$$
or maybe
$$
\int_{5d} w_2Y=0 \mod 2 (?).
$$

*Has this relation anything to do with the Euler class or decomposition of Euler class?

My trial: It looks that if $ Y=Sq^1 w_2 \mod 2$, we have
$$
Sq^1 (w_2^2)= w_2 Y +Y w_2 \mod 2 . 
$$
I dont know does it help to relate the above relation to
$
\int_{5d} Sq^1 (w_2^2)=0 \mod 2 (?).
$ There is an additional factor of 2 for the later equality which makes the later more trivial, but the earlier nontrivial.
 A: There is always a Bockstein connecting homomorphism $\delta:H^n(X;\mathbb{Z}_2)\rightarrow H^{n+1}(X;\mathbb{Z})$ induced by the short exact sequence of coefficients $\mathbb{Z}\xrightarrow{\times 2} \mathbb{Z}\xrightarrow{red_2} \mathbb{Z}_2$.
The composition $red_2\circ\delta:H^n(X;\mathbb{Z}_2)\rightarrow H^{n+1}(X;\mathbb{Z})\rightarrow H^{n+1}(X;\mathbb{Z}_2)$ coincides with the operation $Sq^1$.
Since $\omega_2\in H^2(BO(n);\mathbb{Z}_2)\cong\mathbb{Z}_2$ is a generator satisfying $Sq^1\omega_2=\omega_3\neq 0$, the class $\delta\omega_2\in H^3(BO(n);\mathbb{Z})$ is non-trivial. Since $H^3(BO(n);\mathbb{Z})\cong Ext(H_2(BO(n);\mathbb{Z})),\mathbb{Z})\cong \mathbb{Z}_2$ by the Universal Coefficient Theorem, it must be that $\delta\omega_2$ is a generator. This is exactly the class you have called $Y$. As noted above, its mod 2 reduction is $Sq^1\omega_2=\omega_3$.
The relation has nothing to do with the Euler class, since it is present already in $BO(3)$ (clearly not in $BO(2)$, however).
And with regards to your last comment, note that $Sq^1(\omega_2^2)=2\,\omega_2\cdot Sq^1\omega_2=2\omega_2\cdot \omega_3=0$, which is the same as $Y\cdot\omega_2+\omega_2\cdot Y=2\,Y\cdot\omega_2\equiv 0\mod 2$.
Edit:  I'm afraid I don't know what your $\int_{5d}\omega_2 Y$ notation means. If you explain it, maybe I can edit in the details you are looking for.
Since you are reducing mod 2, then why not use $red_2(Y)=\omega_3$? Then you are looking for the evaluation
$$\langle \omega_2\omega_3,[M^5]\rangle\in\mathbb{Z}_2$$
of $\omega_2\omega_3$ against the fundamental class of a given 5-manifold $M=M^5$. According to Wikipedia this particular Stiefel-Whitney number is called the de Rham Invariant of $M$ (https://en.wikipedia.org/wiki/De_Rham_invariant). There are other ways to characterise this invariant, and you can read about them in the paper "Semi-Characteristics and  Cobordism "by Lusztig, Milnor and Peterson, (available here https://core.ac.uk/download/pdf/82791020.pdf).
I haven't read their paper completely, so don't know if they give any 5-dimensional examples, but I managed to find the Wu Manifold 
$$M^5=SU(3)/SO(3).$$
This is an orientable simply-connected 5-manifold with
$$H^*(M;\mathbb{Z_2})=\mathbb{Z}_2[x_2,x_3]/(x_2^2,x_3^2),\qquad Sq^1x_2=x_3,$$
where $x_i\in H^i(M;\mathbb{Z}_2)$. We have $\omega_2(M)=x_2$ and therefore $\omega_3(M)=Sq^1\omega_2(M)=x_3$. Since $x_2x_3$ generates $H^5(M;\mathbb{Z}_2)$ we have
$$\langle \omega_2\omega_3,[M^5]\rangle\neq 0.$$
