Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We have a parallel plate capacitor that has a net charge of $\pm$400 nC when the potential difference between its plate is 100 V. What is its capacitance?
And if we know that the plates' dimensions are 10 cm$\times$ 10 cm, what is the separation distance between the two plates?
$$\color{blue}{\bf [b]}$$
Solving the first given formula for $C$,
$$C=\dfrac{400\times 10^{-9}}{100}=\color{red}{\bf 4}\;\rm nF$$
Solving the second given formula for $d$,
$$d=\dfrac{(8.85\times 10^{-12})(0.1\times 0.1)}{4\times 10^{-9}}=\color{red}{\bf 22.1}\;\rm \mu m$$
