The charge accumulated on the capacitor connected in the following...

Under the balanced condition, the capacitor will act as an open circuit.

The current $\left({i}_{1}\right)$ through the $1Ω$ resistor is given by

$\begin{cases}{i}_{1} & = & \dfrac{10V}{1Ω + 2Ω} \\ & = & \dfrac{10}{3}A\end{cases}$

The current $\left({i}_{2}\right)$ through the $6Ω$ resistor is given by

$\begin{cases}{i}_{2} & = & \dfrac{10V}{6Ω + 4Ω} \\ & = & 1A\end{cases}$

From the above diagram, it can be written that

${V}_{A} + {i}_{1} \times 1 - {i}_{2} \times 6 = {V}_{B} ⇒ {V}_{A} - {V}_{B} = 1 \times 6 - \dfrac{10}{3} \times 1 = \dfrac{8}{3}V$

Hence, the charge on the capacitor is given by

$\begin{cases}Q & = & C\left({V}_{A} - {V}_{B}\right) \\ & = & 150\mu F \times \dfrac{8}{3}V \\ & = & 400\mu C\end{cases}$