In the circuit shown below, the charge on the 60μ F capacitor is

The circuit shown in the figure is a balanced Wheatstone bridge, so the $80\mu F$ doesn't have any effect on the circuit and it can be removed.

After removing the $80\mu F$, we see that the $30\mu F$ and $60\mu F$ are in series. Similarly, the $15\mu F$ and $30\mu F$ are also in series.

${C}_{1} = \dfrac{30 \times 60}{30 + 60} = 20\mu F$, ${C}_{2} = \dfrac{15 \times 30}{15 + 30} = 10\mu F$

Hence, the equivalent capacitance of the entire circuit is

$C = {C}_{1} + {C}_{2} = 30\mu F$

${C}_{eq} = \dfrac{C \times 30}{C + 30} = 15\mu F$

The charge that flows through the battery = $10 \times 15 = 150\mu C$

This charge will be divided into two branches in the ratio of capacitances.

On solving we get that charge on $60\mu F=100\mu C$