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?

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.
