= Solution
The upper positive plate gives a downward field $E=V/a$, while $\mathbf B$ points into the page. For a rightward electron, the electric force is upward and the magnetic force downward. Its initial speed is
$$
v_0=\sqrt{\frac{2(1\,\mathrm{eV})}{m_e}}=5.931\times10^5\,\mathrm{m\,s^{-1}},\qquad
\frac EB=\frac{20000}{0.0337(1)}=5.935\times10^5\,\mathrm{m\,s^{-1}}.
$$
These are equal to the precision intended by the rounded plate spacing: the arrangement is a <velocity selector>. Accordingly,
$$
\boxed{\theta_{\rm exit}\simeq0.}
$$
It is not appropriate to calculate a large electric deflection and ignore magnetic bending. For completeness, if all printed values are treated literally, let $v_d=E/B$, $u=v_0-v_d$, and $\omega_c=eB/m_e$. The exact nonrelativistic crossed-field motion is
$$
x=v_dt+\frac{u}{\omega_c}\sin\omega_ct,\qquad y=\frac{u}{\omega_c}(\cos\omega_ct-1),
$$
with $v_x=v_d+u\cos\omega_ct$ and $v_y=-u\sin\omega_ct$. At the exit solve $x(t_e)=0.50$ m, then use $\theta=\operatorname{atan2}(v_y,v_x)$. The small rounding mismatch produces nanometre-scale oscillations, with $|\theta|\le\sin^{-1}(|u|/v_d)\simeq6.32\times10^{-4}$ rad. Its precise oscillatory exit angle is not robust to the supplied rounding.
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