When $K_1$ is closed upper circuit is connected to source





$Q=CV$

Energy $U_1=\frac{1}{2}C{V}^2$

When $K_2$ is also closed complete circuit is connected to source



Therefore the Equivalent circuit



$C_{equi}=C+\frac{2C}{3}=\frac{5C}{3}$

$U_f=\frac{1}{2}\frac{5C}{3}{V}^2=\frac{5C}{6}{V}^2$

Charge supplied by battery after closing $K_2$ is

$Q=\frac{5}{3}CV-CV=\frac{2}{3}CV$

Energy supplied by battery $=U_f-U_i+H$

$\frac{2}{}C{V}^2=\frac{5}{6}C{V}^2-\frac{1}{2}C{V}^2+H$

$3$

$H=\frac{1}{3}C{V}^2$