A parallel plate capacitor having capacitance 12pF is charged by a...

Initial energy of capacitor

${U}_{i} = \dfrac{1}{2}\dfrac{{q}^{2}}{c}$
$= \dfrac{1}{2} \times \dfrac{120 \times 120}{12} = 600pJ$

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

${U}_{f} = \dfrac{1}{2}\dfrac{{q}^{2}}{ck}$

$= \dfrac{1}{2} \times \dfrac{120 \times 120}{12 \times 6 . 5} = 92pJ$

$W + {U}_{f} = {U}_{i}$
$W = 508pJ$