Redrawing network, we get a balanced Wheatstone's network.

Charge on capacitor between terminals $a$ and $b$

$\frac{Q}{2}=\frac{CV}{2}$

Energy stored in that capacitor$=\frac{1}{2}\frac{{\left(\frac{Q}{2}\right)}^2}{C}$

$$$=\frac{Q^2}{8C}$

$=\frac{{\mathrm{CV}}^2}{8}$

$=\frac{1\times {10}^{-6}\times {10}^2}{8}\mathrm{J}$

$=12.5\mu \mathrm{J}$