Answer
a) ${\bf 85.9}\;\rm mT$
b) ${\bf1.62\times 10^{-14} }\;\rm J$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We have here charged particles (which are the electrons) that undergo a uniform circular motion at a constant speed due to the perpendicular uniform magnetic field.
Recalling that the frequency of such motion (cyclotron motion) is given by
$$f_{\rm cyc}=\dfrac{qB}{2\pi m}$$
Hence, the magnetic field strength is given by
$$B=\dfrac{2\pi m f_{\rm cyc}}{q} $$
Plug the known;
$$B=\dfrac{2\pi (9.11\times 10^{-31})(2.4\times 10^9)}{(1.6\times 10^{-19})} $$
$$B=\color{red}{\bf 85.9}\;\rm mT$$
$$\color{blue}{\bf [b]}$$
The maximum kinetic energy of the electron occurs when it is at maximum velocity just before hits the wall.
$$K_{\rm max}=\frac{1}{2}mv_{\rm max}^2$$
where $v=2\pi r_{\rm max}/T$ where $T=1/f$, so that $v=2\pi r_{\rm max} f$.
$$K_{\rm max}=\frac{1}{2}m(2\pi r_{\rm max} f)^2$$
$$K_{\rm max}=2\pi ^2 m r_{\rm max}^2 f^2$$
Plug the known;
$$K_{\rm max}=2\pi ^2 (9.11\times 10^{-31}) (1.25\times 10^{-2})^2 (2.4\times 10^9)^2$$
$$K_{\rm max}= \color{red}{\bf1.62\times 10^{-14} }\;\rm J$$
