In the arrangement shown, energy stored in $3\mu F$ capacitor is 

$C_e=\frac{3\times 6}{3+6}\mu C=2\mu C$

Charge on each capacitor, $Q=C_e.V=2\times 10\mu C=20\mu C$

$\therefore$ Energy stored in $3\mu F$ capacitor, $U=\frac{Q^2}{2C}=\frac{{20}^2}{2\times 3}\mu J=66.67\mu J$