The total energy stored in the condenser system shown in the figure will be__ $\left(\mu J\right)$

$3\mu F$  and $6\mu F$ in series gives  $\frac{3\times 6}{3+6}=2\mu F$. The two  $2\mu F$ in parallel gives  $4\mu F$.
Total energy stored = energy stored in the equivalent capacitor  
$\frac{1}{2}C{V}^2=\frac{1}{2}\times 4\times 2^2=8\mu J$