
$V_{1}+V_{2}=\frac{3}{3}+\frac{3}{2}=2.5 K V$
$V_{3}+V_{4}=\frac{6}{7}+\frac{6}{3}=\frac{20}{7} K V \Rightarrow E=2.5 K V$
Note: $V_{1,2,3,4}$ are the potential differences across $C_{1,2,3,4}$ and $Q_{1,2,3,4}$ are the final charges stored in $C_{1,2,3,4}$ respectively.

