From about circuit diagram $$C_1$$ & $$C_2$$ and $$C_4$$ & $$C_5$$ are shorted with wire. So total potential drop to move from $$C_1$$ to $$C_2$$ and $$C_4$$ to $$C_5$$ will be zero.
$$\therefore C_{eq} = C_3 = 2 \mu F$$
charge supply by battery
$$q = C_{eq} \times V$$
$$= 2 \times 10^{-6} \times 5$$
$$q = 10 \times 10^{-6}$$
$$q = 10 \mu C$$
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