Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
A proton is moving at a constant acceleration of $2\times 10^{12}\;\rm m/s^2$ under a uniform electric field $E$. This electric field is between two parallel plates, where the dimension of the plate is 20 cm times 20 cm [length and width]. Find the charge $Q$ in each plate.
$$\color{blue}{\bf [b]}$$
We need to solve the first given formula for $E$,
$$E=\dfrac{(2\times 10^{12})(1.67\times 10^{-27})}{(1.6\times 10^{-19})}\tag 1$$
Now we need to solve the second given formula for $Q$,
$$Q=(8.85\times 10^{-12})(0.02)E$$
Plug from (1),
$$Q=(8.85\times 10^{-12})(0.02)^2\dfrac{(2\times 10^{12})(1.67\times 10^{-27})}{(1.6\times 10^{-19})}$$
$$Q=\color{red}{\bf 7.4\times 10^{-11}}\;\rm C$$
