In steady state:

 $R_{eq}=12 \mathrm{\Omega }$

Current through the branches with resistance, $I=\frac{6}{12}=0.5 \mathrm{A}$

Now, potential drop across $C_1=3 \mathrm{V}$

and potential drop $C_2=4 \mathrm{V}$

Charge on the first capacitor, $q_1=C_1 V_1=12 \mu C$

Charge on the second capacitor, $q_2=C_2 V_2=24 \mu C$

Therefore, $\frac{q_1}{q_2}=\frac{1}{2}$