The charge on capacitor of capacitance $15\mu \mathrm{F}$ in the figure given below is

As we know, $Q=CV$

Here all capacitors are connected in series.

Thus, equivalent capacitance is $C_{eq}={\left(\frac{1}{10}+\frac{1}{15}+\frac{1}{20}\right)}^{-1}$ $={\left(\frac{6+4+3}{60}\right)}^{-1}=\frac{60}{13} \mathrm{\mu F}$

Therefore, charge $Q=\frac{60}{13}\times 13=60 \mathrm{\mu C}$.

As the capacitors are connected in series, charge on each capacitor wil be same.