In the given circuit. C 1 = 2 μ F , C 2 = 0 . 2 μ F , C 3 = 2 μ F , C 4 = 4 μ F , C 5 =…

From the diagram of the circuit, $C_3, C_4, C_5$ are in series, so the total capacitance is 

$$\begin{gathered} \frac{1}{C\text{'}}=\frac{1}{2}+\frac{1}{4}+\frac{1}{2} \\ \Rightarrow C\text{'}=\frac{4}{5} \mathrm{\mu F} \end{gathered}$$

$C\text{'} \& C_2$ are parallel, the total capacitance is 

$C\text{'}\text{'}=\left(\frac{4}{5}+0.2\right) \mathrm{\mu F}=1 \mathrm{\mu F}$

$C_1, C\text{'}\text{'}, C_6$ are in series. So the equivalent capacitance is,

$$\begin{gathered} \frac{1}{C_{eq}}=\frac{1}{2}+1+\frac{{\displaystyle 1}}{{\displaystyle 2}} \\ \Rightarrow C_{eq}=0.5 \mathrm{\mu F} \end{gathered}$$

Using $Q=C_{eq}V$.

The charge through the battery, $Q$ $=10\times \frac{1}{2}=5 \mathrm{\mu C}$

 From the diagram above,

$Q\text{'}=\frac{5\mathrm{\mu C}\times 0.8}{0.8+0.2}=4 \mathrm{\mu C}$