The potential drop across $7\mathrm{\mu F}$  capacitor is 6V

 

$\begin{array}{l}{\mathrm{q}}_7=\mathrm{CV}=42\mu C\\ Thus, q_3=42\mu C;\\ {\mathrm{V}}_3=\frac{42}{3}=14V\\ V_{3.9}=20\mathrm{V};\\ {\mathrm{q}}_{3.9}=\mathrm{CV}=78\mu C\\ \mathrm{q}=42+78=120\mathrm{\mu C}\\ V_{12}=\frac{\mathrm{q}}{{\mathrm{C}}_{12}}=\frac{120}{12}=10V\\ E={\mathrm{V}}_{3.9}+{\mathrm{V}}_{12}=30\mathrm{V}\end{array}$