Answer
(a) $C/A = 1.18 \times 10^{-6} \mathrm{~F/cm^2}$
(b) $E = 11.33 \times 10^{5} \,\text{V/m}$
Work Step by Step
(a) The capacitance depends on the area $A$ of the plate and the distance between the two plates and it is given by
$$C = \dfrac{K \epsilon_o A}{d}$$
Substitute by the values of $\epsilon_o$ and $d$ to get the capacitance $C$ per area where $A$ is in $\mathrm{m^2}$
\begin{align*}
C/A &= \dfrac{K \epsilon_o }{d} = \dfrac{(10) (8.85 \times 10^{-12} \,\text{F/m}) } {7.5 \times 10^{-9} \,\text{m}} = 1.18 \times 10^{-2} \mathrm{~F/m^2} = \boxed{1.18 \times 10^{-6} \mathrm{~F/cm^2}}
\end{align*}
(b) The maximum voltage is applied at the maximum electric field where the maximum electric field is given by
$$ E = \dfrac{V}{K d} = \dfrac{(85 \times 10^{-3} \,\text{V})}{10 (7.5 \times 10^{-9} \,\text{m})} = \boxed{ 11.33 \times 10^{5} \,\text{V/m}}$$
