Answer
(a) $V_{ab} = 604 \,\text{V}$
(b) $A = 9 \times 10^{-3} \mathrm{~m^2}$
(c) $E = 1840 \,\text{kV/m}$
(d) $E = 16.3 \times 10^{-6} \mathrm{~C/m^2}$
Work Step by Step
(a) The potential difference between the two plates is given by
$$V_{ab} = \dfrac{Q}{C}$$
Substitute with the values of $Q$ and $C$ to get $V_{ab}$
$$V_{ab} = \dfrac{Q}{C} = \dfrac{0.148 \times 10^{-6}\,\text{C}}{245 \times 10^{-12} \,\text{F}} = \boxed{604 \,\text{V}} $$
(b) The capacitance $C$ of the two plates is given by
\begin{gather*}
C = \dfrac{\epsilon_o A}{d}\\
A = \dfrac{Cd}{\epsilon_o}
\end{gather*}
Substitute with the values of $d, \epsilon_o$ and $C$ to get the area of the plate $A$
\begin{align*}
A = \dfrac{Cd}{\epsilon_o} = \dfrac{(245 \times 10^{-12} \,\text{F})(0.328 \times 10^{-3} \,\text{m})}{8.85 \times 10^{-12} \mathrm{~C^2/N\cdot m^2}} = \boxed{9 \times 10^{-3} \mathrm{~m^2} }
\end{align*}
(c) The electric field $E$ equals the potential difference divided by the distance between the two plates
$$E = \dfrac{V_{ab}}{d} =\dfrac{604 \,\text{V}}{0.328 \times 10^{-3} \,\text{m}} = \boxed{1840 \,\text{kV/m}}$$
(d) The surface charge density $\sigma$ equals the electric field times the permittivity of the free space $\epsilon_o$
$$\sigma = \epsilon_o E = (8.85 \times 10^{-12} \mathrm{~C^2/N\cdot m^2}) (1840 \,\text{kV/m}) = \boxed{16.3 \times 10^{-6} \mathrm{~C/m^2}}$$
