Let $M$ be oriented manifold: this happens if and only if $w_1(M)=0$ (the first Stiefel Whitney class, being an element in $H^1(M,\mathbb{Z}_2)$. There is a result that if $M$ is three dimensional then from the vanishing of $w_1$ automatically follow that also $w_2=0$. This should be rather easy consequence of properties of Wu classes and Steenrod squares however I don't see how to proceed in the calculation. I expressed the total Stiefel Whitney class $w$ of $M$ as $Sq(v)$ where $v$ is the total Wu class. After evaluating $Sq(v)$ I arrived, using the fact that $M$ is 3 dimensional (if I'm not mistaken) at the following: $1+w_1+w_2+w_3=1+v_1+v_2++v_1 \cup v_1$. From $w_1=0$ we infer $v_1=0$ therfore also $v_1 \cup v_1=0$ and the question boils down to showing $v_2=0$ but I don't see how to prove this.