A parallel plate capacitor having capacitance 12   pF is charged by a battery to a potential…

Initial energy of capacitor

$U_i=\frac{1}{2}\frac{q^2}{c}$
$=\frac{1}{2}\times \frac{120\times 120}{12}=600 \mathrm{pJ}$

Since battery is disconnected so charge remain same. Final energy of capacitor

 $U_f=\frac{1}{2}\frac{q^2}{c k}$

$=\frac{1}{2}\times \frac{120\times 120}{12\times 6.5}=92 \mathrm{pJ}$

$W+U_f=U_i$
$W=508 \mathrm{pJ}$