$\newcommand{\Sq}{\operatorname{Sq}}$ For this exercise, it is useful to recall the following property of the Steenrod squares
$$\Sq^k: H^n(M; \Bbb Z/2) \longrightarrow H^{n+k}(M; \Bbb Z/2):$$
$(\ast)$ If $a \in H^n(M)$, then $\Sq^0(a) = a$, $\Sq^n(a) = a \smile a$, and $\Sq^k(a) = 0$ for all $k > n$.
The axiomatic description of the Steenrod squares, which includes the above property as an axiom, is given by Milnor and Stasheff in Chapter 8, where they use the Steenrod squares to construct Stiefel-Whitney classes.
We can use $(\ast)$ to compute all the Stiefel-Whitney classes of a compact, oriented $3$-manifold $M$. In what follows, we write $v_k$ for the degree $k$ part of the total Wu class
$$v = 1 + v_1 + v_2 + v_3.$$
Since $M$ is oriented, we know that $w_1 = 0$. By Wu's formula,
$$w_1 = \Sq^1(1) + \Sq^0(v_1) = v_1,$$
where we used $(\ast)$ twice. Hence $v_1 = 0$ in this case.
Now let us show that $w_3$ is zero. First, by the definition of the Wu class,
$$v_2 \smile x = \Sq^2(x)$$
for all $x \in H^1(M; \Bbb Z/2)$. But since $1 < 2$, we have that
$$v_2 \smile x = 0$$
for all $x \in H^1(M; \Bbb Z/2)$, and hence
$$v_2 = 0.$$
Similarly,
$$v_3 = 0.$$
Now by Wu's formula,
$$w_3 = \Sq^3(1) + \Sq^2(v_1) + \Sq^1(v_2) + \Sq^0(v_3) = 0$$
by using $(\ast)$ on the first two terms and the fact that $v_2 = v_3 = 0$ on the last two terms.
Finally, by Wu's formula,
$$w_2 = \Sq^2(1) + \Sq^1(v_1) + \Sq^0(v_2) = 0$$
by using $(\ast)$ on the first term and the fact that $v_1 = v_2 = 0$ on the last two terms.
We conclude that $w_1 = w_2 = w_3 = 0$.
