Solution :
Area of the plates of a parallel plate capacitor, $A = 90 cm^2 = 90 \times 10^{−4} m^2$
Distance between the plates, $d = 2.5 mm = 2.5 \times 10^{−3}\; m $
Potential difference across the plates, $V = 400\; V $
Capacitance of the capacitor is given by the relation,
$C= \large\frac{\in_0 A}{d}$
Electrostatic energy stored in the capacitor is given by the relation,
$E_1=\large\frac{1}{2}$$CV^2$
$\qquad=\large\frac{1}{2} \frac{\in_0 A}{d}$$v^2$
Where
$\in_0$= Permittivity of free space $= 8.85 × 10^{−12} C ^{2} N ^{−1} m ^{−2}$
$E_1= \large\frac{1 \times 8.85 \times 10^{-12} \times 90 \times (400)^2}{2 \times 2.5 \times 10^{-3}}$$=2.55 \times 10^{-6}\;J$