A galvanometer (G) of 2Ω resistance is connected in the given...

In steady state:

 ${R}_{eq} = 12Ω$

Current through the branches with resistance, $I = \dfrac{6}{12} = 0 . 5A$

Now, potential drop across ${C}_{1} = 3V$

and potential drop ${C}_{2} = 4V$

Charge on the first capacitor, ${q}_{1} = {C}_{1}{V}_{1} = 12\muC$

Charge on the second capacitor, ${q}_{2} = {C}_{2}{V}_{2} = 24\muC$

Therefore, $\dfrac{{q}_{1}}{{q}_{2}} = \dfrac{1}{2}$