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

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

$C_1=\frac{30\times 60}{30+60}=20 \mathrm{\mu F}$, $C_2=\frac{15\times 30}{15+30}=10 \mathrm{\mu F}$

Hence, the equivalent capacitance of the entire circuit is

 $C=C_1+C_2=30 \mathrm{\mu F}$

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

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

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

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