The potential drop across capacitor is 6V. Then

Potential drop across 7µf is 6V
$\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}=\frac{{\mathrm{C}}_2}{{\mathrm{C}}_1}=\frac{3}{7};\frac{6}{{\mathrm{V}}_2}=\frac{3}{7}\Rightarrow {\mathrm{V}}_2=14\mathrm{V}$
V₁ + V₂ = 6 + 14 = 20V
7, 3 are in series $\Rightarrow \frac{7\times 3}{10}=2.1\mathrm{\mu f}$
2. 1, 3. 9 are in parallel
$\Rightarrow$ 2.1+ 3.9 = 6µf
12, 6 are in series so

12, 6 are in series so
$\begin{array}{l}\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}=\frac{{\mathrm{C}}_2}{{\mathrm{C}}_1}=\frac{6}{12}=\frac{1}{2}\\ \frac{{\mathrm{V}}_1}{20}=\frac{1}{2}\Rightarrow {\mathrm{V}}_1=10\mathrm{V}\end{array}$
${\mathrm{V}}_{\mathrm{net}}={\mathrm{V}}_1+{\mathrm{V}}_2=10+20=30\mathrm{V}$