A capacitor of capacitance C 1 = 1   μ F is charged using a 9   V battery. C 1 is then…

Initially, when $C_1$ is connected with $9 \mathrm{V}$ the charge on it will be,

$q_0=C_{1}V=9 \mathrm{\mu C}$.

Later all the capacitors are connected in parallel. Therefore, the potential difference between the capacitors will be equal. If the charges on them are $q_1, q_2 \text{and} q_3$. Then by conservation of charge,

$q_1+q_2+q_3=q_0=9 \mathrm{\mu C} \cdots \left(1\right)$

And

$$\begin{gathered} \frac{q_1}{C_1}=\frac{q_2}{C_2}=\frac{q_3}{C_3} \\ \Rightarrow \frac{{\displaystyle q_1}}{{\displaystyle 1}}=\frac{{\displaystyle q_2}}{{\displaystyle 2}}=\frac{{\displaystyle q_3}}{{\displaystyle 3}} \cdots \left(2\right) \end{gathered}$$

Using the above two equations we get $q_3=4.5 \mathrm{\mu C}=4.5\times {10}^{-6} \mathrm{C}$.