From the diagram of the circuit, C3,C4,C5 are in series, so the total capacitance is
C′1=21+41+21⇒C′=54μF
C'&{C}_{2} are parallel, the total capacitance is
C′′=(54+0.2)μF=1μF
C1,C′′,C6 are in series. So the equivalent capacitance is,
Ceq1=21+1+21⇒Ceq=0.5μF
Using Q=CeqV.

The charge through the battery, Q =10×21=5μC
From the diagram above,
Q′=0.8+0.25μC×0.8=4μC