The charge on capacitor of capacitance 15 in the figure given below...

As we know, $Q = CV$

Here all capacitors are connected in series.

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

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

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