$i=\frac{9}{27 \times 10^{3}}$

The voltage is given by,

$V_{c}=i \times 15 \times 10^{3}$

The initial charge is given by,

$q_{0}=5 \times 10^{-6} \times \frac{9}{27} \times 15$

$=25 \mu C$

The charge when switch is open is given by,

$q=q_{0} e^{\frac{-t}{R C}}$

Substitute $25 \mu C$ for $q_{0}, 1$ sec for $t, 20 \times 10^{3}$ for $R$ and $5 \times 10^{-6}$ for $C$ in equation $(I).$

$q=(25 \mu C ) e^{\frac{-(1 \sec )}{\left(20 \times 10^{3}\right)\left(5 \times 10^{-6}\right)}}$

$=25 e^{-10} \mu C$