In the circuit shown (figure), if switches $S_{1}and{S}_2$ have been closed for along time, then the charge on the capacitor 

After long time, current through the capacitor = 0

Therefore, current through the  $6\Omega$ resistor, $i=\frac{12-4}{8}=1A$

Voltage across capacitor,  $V=4+\left(6\right)\left(1\right)=10V$

Charge on the capacitor,  $Q=\left(10\right)\left(10\right)=100\mu C$

After the insertion of the dielectric, we get

$C\text{'}=\frac{A{\epsilon }_0}{\frac{d}{3}+\frac{d}{3}+\frac{d}{3\left(2\right)}}=\frac{A{\epsilon }_0}{d}\left[\frac{1}{(\frac{1}{3}+\frac{1}{3}+\frac{1}{6})}\right]$

$=\left(10\right)\left(\frac{1}{(5/6)}\right)=\frac{5}{6}\left(10\right)=12\mu F$

Hence, voltage across the capacitor remains the same

Charge on the capacitor, $Q\text{'}=\left(12\right)\left(10\right)=120\mu C$