Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the current is given by
$$I=\dfrac{Q}{t}$$
where $Q=Ne$ where $N$ is the number of electrons and $e$ is the charge of one electron.
$$I=\dfrac{Ne}{t}$$
So the number of electrons that hit the screen each second is given by
$$N=\dfrac{tI}{e}$$
Plug the known;
$$N=\dfrac{(1)(50\times 10^{-6})}{1.6\times 10^{-19}}$$
$$N=\color{red}{\bf 3.125\times 10^{14}}\;\rm electron$$
$$\color{blue}{\bf [b]}$$
We know that the current density is given by
$$J =\dfrac{I}{A}=\dfrac{I}{\pi r^2}$$
Plug the known;
$$J =\dfrac{(50\times 10^{-6})}{\pi (0.20\times 10^{-3})^2}$$
$$J =\color{red}{\bf 398}\;\rm A/m^2$$
$$\color{blue}{\bf [c]}$$
To find the electric force exerted on the electron, we need to find the force exerted on it. And to find the force exerted on the electron, we need to find its acceleration.
$$v_f^2=v_i^2+2ad$$
where $v_i=0$ since it starts from rest.
Hence,
$$a=\dfrac{v_f^2}{2d}\tag 1$$
The electric force exerted on the electron is given by
$$F=eE=m_ea$$
Hence,
$$E=\dfrac{m_ea}{e}$$
Plug from (1),
$$E=\dfrac{m_ev_f^2}{2ed}$$
Plug the known;
$$E=\dfrac{(9.11\times 10^{-31})(4\times 10^7)^2}{2(1.6\times 10^{-19})(5\times 10^{-3})}$$
$$E=\color{red}{\bf9.11\times 10^5}\;\rm V/m$$
$$\color{blue}{\bf [d]}$$
We know that the power here is the rate at which the screen absorbs the energy of the electrons.
Hence, the net power due to the electrons' beam is given by
$$P=\dfrac{N\Delta E}{\Delta t}$$
where $\Delta E=\Delta K =|\frac{1}{2}mv_{f2}^2-\frac{1}{2}mv_f^2|=0- \frac{1}{2}mv_i^2$
where $v_{f2}=0 $ since the electron stops after hitting the screen.
$$P=\frac{1}{2}m_ev_f^2\dfrac{ N }{\Delta t}$$
Plug the known and recall that we found the number of electrons that hit the screen each second ($N/\Delta t$) in part (a) above.
$$P=\frac{1}{2} (9.11\times 10^{-31}) (4\times 10^7)^2 (3.125\times 10^{14}) $$
$$P=\color{red}{\bf 0.23}\;\rm W$$
