Given: $\mathrm{v}=5.20 \times 10^6 \mathrm{~m} / \mathrm{s}, \mathrm{B}=1.3 \times 10^{-4} \mathrm{~T}$, $\frac{\mathrm{e}}{\mathrm{m}}=1.76 \times 10^{11} \mathrm{C} / \mathrm{kg}$.
(a) When magnetic field is applied perpendicular (normal) to a moving $\mathrm{e}^{-}$, it moves in a circular path whose radius is given by $B e v=\frac{m v^2}{r}$ or $r=\frac{m v}{B e}$ $=\frac{\mathrm{v}}{\mathrm{Be} / \mathrm{m}}=\frac{5.2 \times 10^6}{1.3 \times 10^{-4} \times 1.76 \times 10^{11}}$ $=0.2273 \mathrm{~m}$.
(b) For an electron beam of energy $20 \mathrm{MeV}$, mass has to be calculated using relativistic picture, hence
$$
\begin{aligned}
& m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}} \\
\Rightarrow r &=\frac{v}{B\left(\frac{e}{m_0}\right) \sqrt{1-v^2 / c^2}}
\end{aligned}
$$
Total energy $E=m^2=\frac{m_0 c^2}{\sqrt{1-\frac{v^2}{c^2}}}$
$$
\begin{aligned}
&\Rightarrow \sqrt{1-\frac{\mathrm{v}^2}{\mathrm{c}^2}}=\frac{\mathrm{m}_0 \mathrm{c}^2}{\mathrm{E}} \\
&\Rightarrow \sqrt{1-\frac{\mathrm{v}^2}{\mathrm{c}^2}}=\frac{0.51 \mathrm{MeV}}{20 \mathrm{MeV}}=0.0255 \\
&\Rightarrow 1-\frac{\mathrm{v}^2}{\mathrm{c}^2}=(0.0255)^2=0.00065
\end{aligned}
$$
$\begin{aligned} & \Rightarrow \frac{\mathrm{v}^2}{\mathrm{c}^2}=(1-0.00065)=0.99935 \\ & \Rightarrow \mathrm{v}=0.9997 \mathrm{c} \text { substitute in eqs. (1) } \\ \therefore \quad r &=\frac{0.9997 \times 3 \times 10^8}{1.3 \times 10^{-4} \times 1.76 \times 10^{11} \times 0.0255} \\ &=514.04 \mathrm{~m} \end{aligned}$