Solution :
Capacitance of the capacitor, C = 600 pF
Potential difference, V = 200 V
Electrostatic energy stored in the capacitor is given by,
$E_1= \large\frac{1}{2} CV^2 =\large\frac{1}{2} \times (600 \times 10^{-12}) \times (200)^2j = 1.2 \times 10^{-5}$
If supply is disconnected from the capacitor and another capacitor of capacitance $C = 600\; pF$ is connected to it, then equivalent capacitance $(C_{eq})$ of the combination is given by,
$\large\frac{1}{C_{eq}}=\frac{1}{C}+\frac{1}{C}$
=> $\large\frac{1}{C_{eq}}=\frac{1}{600}+\frac{1}{600} =\frac{2}{600}$
=> $C_{eq} =300\;pF$
New electrostatic energy can be calculated as
$E_2= \large\frac{1}{2} $$ \times 300 \times (200)^3\;J =0.6 \times 10^{-5}\;J$
Answer : $ 6 \times 10^{-6}\;J$