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. 



Vanishing of $w_2$ for orientable 3-manifolds
From the relation $Sq(v)=w$ we get that $v\_2=w\_2+w\_1^2$. More specifically By Thm 11.14 (which is just an extended form of the above equality) in M-S you have $w\_1=v\_1$ and $w\_2=Sq^1(v\_1)+v\_2=v\_1^2+v\_2$. Plug them together you obtain my first claim. 


Recall we have a nice characterisation of the (second) Wu-class, as $$Sq^{2}(x)=v\_2\smile x$$ for any $x \in H^{n-k}(M;\Bbb Z\_2)$ (See Milnor Stasheff page 132). Since in our case $n=3$, we get that for any $x\in H^1(M;\Bbb Z\_2)$, $0=v\_2\smile x$, which gives you $v\_2=0$. 


To see this last equality, assume $v\_2 \neq 0$. We would have $$1=\langle v\_2, (v\_2)\_\*\rangle=\langle v\_2, [M]\frown x\rangle=0$$ Where $(v\_2)\_\* \in H\_2(M;\Bbb Z\_2)$ is the dual of $v\_2$ and Poincaré Duality (mod 2) tells us that there exists an element $x \in H^1(M; \Bbb Z\_2)$ such that $[M]\frown x=(v\_2)\_\*$


Therefore you have $0=w\_2+w\_1^2$, which proves your claim


