Answer
$A =0.63 m^{2}$ should be the minimum area of the plates of capacitor.
Work Step by Step
The capacitance is given by $C= kC_{0} = Kε_{0}\frac{A}{d}$ where $C_{0}$ is capacitance without the dielectric, k is the dielectric constant, A is the plate area, and d is the plate separation. The electric field between the plates is given by
$E = \frac{V}{d}$ where V is the potential difference between the plates. Thus, $d = \frac{V}{E}$ and $C= kε_{0}\frac{AE}{V}$
=>$A= \frac{CV}{kε_{0}E}$
For the area to be a minimum, the electric field must be the greatest it can be without breakdown occurring as,
$A= \frac{(7 \times 10^{-8})(4 \times 10^{3})}{2.8(8.85 \times 10^{-12})(18 \times 10^{6})}$
$A =0.63 m^{2}$
