In the circuit shown (figure), if switches S₁, and S₂ have been closed for a long time, then the charge on the capacitor

After long time, current through the capacitor - 0.
Therefore, current through the 6Ω  resistor  $\mathrm{i}=\frac{12-4}{8}=1\mathrm{A}$
Voltage across capacitor, V - 4 + (6)(1) = 10 V
Charge on the capacitor, Q: (10)(10) = 100 µC
After the insertion of the dielectric, we get
$\begin{array}{r}{\mathrm{C}}^{\mathrm{\prime }}=\frac{{\mathrm{A\epsilon }}_0}{\frac{\mathrm{d}}{3}+\frac{\mathrm{d}}{3}+\frac{\mathrm{d}}{3\left(2\right)}}=\frac{{\mathrm{A\epsilon }}_0}{\mathrm{d}}\left[\frac{1}{\left(\frac{1}{3}+\frac{1}{3}+\frac{1}{6}\right)}\right]\\ =\left(10\right)\left(\frac{1}{(5/6)}\right)=\frac{5}{6}\left(10\right)=12\mathrm{\mu F}\end{array}$
Hence, voltage across the capacitor remains the same.
Charge on the capacitary, Q' : (12) (10) = 120 µC