3 capacitors connected with a 6V battery as shown in the figure below
$\therefore $ Equivalent capacitance of $1\,nf$ and $2\,nF$ connected in parallel is given as, $C'=(1+2) nF$
This equivalent $C'$ is connected in series with $3\,nF$. So, the equivalent capacitance of the given circuit is
$C_{e q}=\frac{(1 \times 2) \times 3}{(1+2)+3}=1.5\,nF$
Total charge on the capacitors $=V \times C_{e q}$
$=6 \times(1.5\,nF )=9\,nC$
