Answer
See the detailed answer below.
Work Step by Step
We know that the magnetic force exerted on a moving charge is given by
$$F=q\vec v\times \vec B\tag 1$$
The velocity of the electron has two components one gives the electron the spiral motion (the circular motion) and the other gives it the linear motion in the positive $z$-direction, as we see in the given figure.
So, from the given figure
$$v_z= v \sin30^\circ \tag 2$$
and
$$v_r=v\cos30^\circ\tag 3$$
Hence, for the electron’s spiral trajectory
$$F=\dfrac{mv_r^2}{r}$$
Plug $F$ from (1),
$$qv_rB=\dfrac{mv_r^2}{r} $$
Solving for $r$ ; where $q=e$
$$r=\dfrac{mv_r^2}{ev_rB}=\dfrac{mv_r}{e B}$$
Plug $v_r$ from (3),
$$r =\dfrac{m v\cos30^\circ}{e B}$$
Plug the known;
$$r =\dfrac{(9.11\times 10^{-31})(5\times 10^6)\cos30^\circ}{(1.6\times 10^{-19})(30\times 10^{-3})}$$
$$r=\color{red}{\bf 0.82}\;\rm mm$$
Recalling that the pitch $P$ is the linear distance traveled in $T$ where $T$ is the period time for one full revolution around $z$-axis.
$$p=v_zT=v_z\dfrac{2\pi r}{v_r}$$
Plug from (2) and (3),
$$p =v \sin30^\circ \dfrac{2\pi r}{v \cos30^\circ }=2\pi r\tan 30^\circ$$
Plug the known;
$$p = 2\pi (0.82)\tan 30^\circ$$
$$p=\color{red}{\bf 2.98}\;\rm mm$$
