Answer
(a) $F = 1.76 \times 10^{-16} \mathrm{N}$
(b) $a = 1.93 \times 10^{14} \mathrm{m} / \mathrm{s}^{2}$
Work Step by Step
(a) First, let us find the electric field due to the potential difference by
$$E=\frac{V}{d}=\frac{22.0 \mathrm{V}}{0.02 \mathrm{m}}=1.10 \times 10^{3} \mathrm{N} / \mathrm{C}$$
Then, the force exerted on charge $q$ will be
$$F=|q| E=\left(1.6 \times 10^{-19} \mathrm{C}\right)\left(1.10 \times 10^{3} \mathrm{N} / \mathrm{C}\right)=\boxed{1.76 \times 10^{-16} \mathrm{N}}$$
(b) From the force value and using Newton's first law we could get the acceleration by
\begin{align}
a&=\frac{F}{m}\\
&=\frac{1.76 \times 10^{-16} \mathrm{N}}{9.1 \times 10^{-31} \mathrm{kg}}\\
&=\boxed{1.93 \times 10^{14} \mathrm{m} / \mathrm{s}^{2}}
\end{align}
