Given that, $C_1=5 \mu \mathrm{F}, C_2=C_3=10 \mu \mathrm{F}$ and $E=20 \mathrm{~V}$
When the switch is connected to point $A$.
Then, situation of the circuit is shown below.


Charge stored in $C_1$,
$$
Q=C_1 V_1=C_1 E=5 \times 20 \mu \mathrm{C}=100 \mu \mathrm{C}
$$
When switch is connect to point $B$, then situation circuit is shown below.

The charge $Q_1$ will start to distribute through, $C_2$ and $C_3$ to make the potential difference $V$ across all the three capacitors $C_1, C_2$ and $C_3$.
Now, common potential difference,
$$
V=\frac{\text { Total charge }}{\text { Total capacity }}=\frac{C_1 V_1+C_2 V_2+C_3 V_3}{C_1+C_2+C_3}
$$
where, $V_1=20 \mathrm{~V}$, but $V_2=V_3=0 \mathrm{~V}$
$$
\therefore \quad V=\frac{100}{5+10+10}=\frac{100}{25}=4 \mathrm{~V}
$$
Now, charge on $C_3, q_3=C_3 V$
$$
\begin{aligned}
& =10 \times 4 \mu \mathrm{C} \\
& =40 \mu \mathrm{C}
\end{aligned}
$$