In the circuit shown in the figure, each capacitor has a...

Let the capacitance of each capacitor be $C = 5.0 \mu\text{F}$ and the voltage of the battery be $V = 50 \text{ V}$.

Before the switch $S$ is closed, the circuit consists of two capacitors in parallel, which are in series with a third capacitor.
The equivalent capacitance of the parallel combination is:
$C_{p} = C + C = 5.0 + 5.0 = 10.0 \mu\text{F}$

This combination is in series with the third capacitor $C$. The initial equivalent capacitance of the entire circuit is:
$C_{initial} = \dfrac{C_{p} \times C}{C_{p} + C} = \dfrac{10.0 \times 5.0}{10.0 + 5.0} = \dfrac{50}{15} = \dfrac{10}{3} \mu\text{F}$

The initial charge supplied by the battery is:
$Q_{initial} = C_{initial} \times V = \dfrac{10}{3} \times 50 = \dfrac{500}{3} \mu\text{C}$

When the switch $S$ is closed, it short-circuits the third capacitor. The circuit now consists only of the parallel combination of the first two capacitors connected directly across the battery.
The final equivalent capacitance is:
$C_{final} = C_{p} = 10.0 \mu\text{F}$

The final charge supplied by the battery is:
$Q_{final} = C_{final} \times V = 10.0 \times 50 = 500 \mu\text{C}$

The amount of charge that flows through the wire $AB$ is the additional charge drawn from the battery:
$\Delta Q = Q_{final} - Q_{initial} = 500 - \dfrac{500}{3} = \dfrac{1000}{3} \mu\text{C}$

$\Delta Q \approx 333.3 \mu\text{C} = 3.33 \times 10^{-4} \text{ C}$