In the given circuit, the initial charges on the capacitors are shown in the figure. The charge flown through the switches $S_{1}and{S}_2,$respectively after closing the switches are

Let $q_1$ charge flows through  $S_{1}and{q}_{2}through{S}_2.$For capacitor of capacitance 2C, 

$Q_0+q_1=2CV=Q_{0}or{q}_1=0$ 

OR 

voltage across 2C= $\frac{Q_0}{2C}=V$

Therefore no charge flow through it.

According KVL, $V-\frac{q_2}{C}=0$

$q_2=CV=C\frac{Q_0}{2C}=\frac{Q_0}{2}$

Change of charge=$\frac{Q_0}{2}-\frac{Q_0}{3}=\frac{Q_0}{6}$

OR

For capacitor of capacitance C, 

$\frac{Q_0}{3}+q_2=CV=\frac{Q_0}{2}or{q}_2=\frac{Q_0}{6}$