Answer
(a) $C = 90 \times 10^{-12} \,\text{F}$
(b) $A = 15 \times 10^{-3} \mathrm{~m^2}$
(c) $V = 4.5 \times 10^6 \,\text{V}$
(d) $U = 1.8 \times 10^{-6} \,\text{J}$
Work Step by Step
(a) The potential between the two plates of the capacitor is related to the charges on the plates and the capacitance and we could get the capacitance by
$$C = \dfrac{Q}{V} = \dfrac{0.0180 \times 10^{-6} \,\text{C}}{200\,\text{V}} = \boxed{ 90 \times 10^{-12} \,\text{F}}$$
(b) The area is related to the capacitance by
\begin{equation}
A = \dfrac{ Cd}{\epsilon_o} \tag{1}
\end{equation}
Let us substitute the values of $\epsilon_o, d$ and $C$ into equation (1) to get the value of $A$
\begin{align}
A &= \dfrac{ Cd}{\epsilon_o} \\
& = \dfrac{ (90 \times 10^{-12} \,\text{F})(0.0015\,\text{m})}{8.85 \times 10^{-12} \,\text{F/m}}\\
&= \boxed{15 \times 10^{-3} \mathrm{~m^2}}
\end{align}
(c) The maximum voltage is applied at the maximum electric field where the maximum electric field is the electric field strength and given by
$$ V = E d = (3 \times 10^6 \,\text{V/m}) (0.0015\,\text{m}) = \boxed{ 4.5 \times 10^6 \,\text{V}}$$
(d) Energy $U$ is the amount of work stored in the capacitor and it is related to the charges on the plates by
$$U = \dfrac{1}{2} QV = \dfrac{1}{2} (0.0180 \times 10^{-6} \,\text{C})(200 \,\text{V}) = \boxed{1.8 \times 10^{-6} \,\text{J}}$$
