b
When the time \(t=\infty,\) the current is,

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

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 \((I).\)

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

\(=25 e^{-10} \mu C\)