The figure shows a capacitor of capacitance C connected to a...

${Q}_{0} = Cε$

$\dfrac{{Q}_{1}}{C} = \dfrac{{Q}_{2}}{3C}{Q}_{1} + {Q}_{2} = {Q}_{0}$

$⇒ {Q}_{1} = \dfrac{{Q}_{0}}{4}; {Q}_{2} = \dfrac{3{Q}_{0}}{4}$

Energy dissipated,
$E = \dfrac{1}{2}{Q}_{0}^{2}C - \dfrac{1}{2}{Q}_{1}^{2}C - \dfrac{1}{2}{Q}_{2}^{2}3C$

$= \dfrac{1}{2C}\left({Q}_{0}^{2} - {Q}_{0}^{2}16 - \dfrac{9{Q}_{0}^{2}}{3 \times 16}\right)$

$= {Q}_{0}^{2}32C\left(16 - 1 - 3\right) = \dfrac{3}{8}{Q}_{0}^{2}C$