Answer
${\bf 45.6}\;\rm mT$
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,
$$B=\dfrac{\Delta V}{vd}\tag 1$$
Now we need to find $v$ which is given by the radius of deflection
$$v=\dfrac{qBr}{m } $$
$$v=\dfrac{eBr}{m_p } $$
Plug into (1),
$$B=\dfrac{m_p \Delta V}{ eBrd} $$
Thus,
$$B^2=\dfrac{m_p \Delta V}{ e rd} $$
$$B =\sqrt{\dfrac{m_p \Delta V}{ e rd} }\tag 2$$
Now we need to find the angle of deflection from the geometry of the figure below.
$$\sin\theta= \dfrac{L}{r} $$
So, $$r=\dfrac{L}{\sin\theta}\tag 3$$
where $\theta=\tan^{-1}\left[\dfrac{\frac{1}{2}d}{L}\right]=\tan^{-1}\left[\dfrac{0.5\times 10^{-2}}{5\times 10^{-2}}\right]=\bf 5.71059^\circ$
Plug into (3);
So,
$$r=\dfrac{L}{\sin 5.71059^\circ} $$
Plug into (2);
$$B =\sqrt{\dfrac{m_p \Delta V\sin 5.71059^\circ}{ e Ld} } $$
Plug the known;
$$B =\sqrt{\dfrac{(1.67\times 10^{-27})(500)\sin 5.71059^\circ}{ (1.6\times 10^{-19})(0.05)(0.5\times 10^{-2})} } $$
$$B=\color{red}{\bf 45.6}\;\rm mT$$