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

The current $\left(i_1\right)$ through the $1 \mathrm{\Omega }$ resistor is given by

$\begin{array}{rcl}{i}_1& =& \frac{10 \mathrm{V}}{1 \mathrm{\Omega }+2 \mathrm{\Omega }}\\ & =& \frac{10}{3} \mathrm{A}\end{array}$

The current $\left(i_2\right)$ through the $6 \mathrm{\Omega }$ resistor is given by

$\begin{array}{rcl}{i}_2& =& \frac{10 \mathrm{V}}{6 \mathrm{\Omega }+4 \mathrm{\Omega }}\\ & =& 1 \mathrm{A}\end{array}$

From the above diagram, it can be written that

$$\begin{gathered} V_A+i_1\times 1-i_2\times 6=V_B \\ \Rightarrow V_A-V_B=1\times 6-\frac{10}{3}\times 1 \\ =\frac{8}{3} \mathrm{V} \end{gathered}$$

Hence, the charge on the capacitor is given by

$\begin{array}{rcl}Q& =& C\left(V_A-V_B\right)\\ & =& 150 \mathrm{\mu F}\times \frac{8}{3} \mathrm{V}\\ & =& 400 \mathrm{\mu C}\end{array}$