The potential drop across 7µF capacitor is 6V. Then

C_eq = 4$\mathrm{\mu }$F ; Charge on 7 $\mathrm{\mu }$F = 6 × 7 = 42$\mathrm{\mu }$C

charge on 3$\mathrm{\mu }$F is same.

Hence pd across it $=\frac{42}{3} , \therefore \boxed{{\mathrm{V}}_2=14\mathrm{V}}$

Hence pd across 3.9$\mathrm{\mu }$F is = 6 + 14 = 20V.

$20=\frac{12}{12+6}\times \mathrm{V}\Rightarrow \mathrm{V}=30\mathrm{V}$