What is the energy stored in the capacitor between terminals a and...

Network is redrawn as shown in figure.
This is a balanced Wheatstone's network. Equivalent capacitance $C_{e q}=C$ Charge on capacitor between the terminals $a$ and $b$ $\frac{Q}{2}=\frac{C V}{2}$ Energy stored in that capacitor
$=\frac{1\left(\frac{Q}{2}\right)^2}{2 C}=\frac{Q^2}{8 C}=\frac{C^2 V^2}{8 C}=\frac{C V^2}{8}$ Given: $C=1 \mu \mathrm{F}, V=10 \mathrm{~V}$ $$\begin{aligned} & =\frac{\left(1 \times 10^{-6}\right) \times 10^2}{8} \\ & =\frac{100}{8} \times 10^{-6}=12.5 \mu \mathrm{J}\end{aligned}$$