Answer
$8.416^\circ$
Work Step by Step
We have here two fields that force the electron beam to move in a straight line. The electric field pushes the electron downward while the magnetic field pushes the electron upward or vice versa.
This means that the two forces exerted by the two fields are equal in magnitude.
Thus,
$$F_B=F_E$$
$$qvB\sin90^\circ=qE$$
Thus,
$$v=\dfrac{E}{B}=\dfrac{\Delta V}{Bd}$$
Plug the known;
$$v =\dfrac{(150)}{(1.0\times 10^{-3})(5\times 10^{-3})}=\bf 3.0\times 10^7\;\rm m/s$$
Now we have a zero potential between the two plates, so the magnetic field deflects the electron into a circular path that has a radius of
$$r=\dfrac{mv}{qB} $$
$$r=\dfrac{m_ev}{eB}=\dfrac{(9.11\times 10^{-31})(3\times 10^7)}{(1.6\times 10^{-19})(1\times 10^{-3})}$$
$$r=\bf 0.170813\;\rm m$$
Now we can find the angle of deflection from the geometry of the figure below.
$$\theta=\sin^{-1}\left[\dfrac{L}{r}\right]$$
Plug the known;
$$\theta=\sin^{-1}\left[\dfrac{(2.5\times 10^{-2})}{(0.170813)}\right]=\color{red}{\bf 8.416}^\circ$$
