The switch S illustrated in the figure is kept closed for a long...

When the switch $S$ is closed for a long time, the circuit reaches a steady state. In this state, the capacitor acts as an open circuit and no current flows through its branch.

The steady current $I$ flows through the outer loop containing the $12\ \text{V}$ battery and the two $10\ \Omega$ resistors in series:
$I = \dfrac{12}{10 + 10} = 0.6\ \text{A}$

The potential difference across the middle $10\ \Omega$ resistor is:
$V_0 = I \times 10 = 0.6 \times 10 = 6\ \text{V}$
Since the capacitor is in parallel with this resistor, the initial voltage across the capacitor is $V_0 = 6\ \text{V}$.

When the switch $S$ is opened at $t = 0$, the battery and the bottom resistor are disconnected. The capacitor now discharges through the middle $10\ \Omega$ resistor.

The time constant $\tau$ of this discharging circuit is:
$\tau = RC = 10 \times 25 \times 10^{-6} = 250\ \mu\text{s} = 0.25\ \text{ms}$

The current $i(t)$ in the discharging circuit at time $t$ is given by:
$i(t) = \dfrac{V_0}{R} e^{-t/\tau}$

Substituting $t = 1.0\ \text{ms}$:
$i(1.0\ \text{ms}) = \dfrac{6}{10} e^{-1.0/0.25} = 0.6 e^{-4}$

Using the value $e^{-4} \approx 0.0183$:
$i(1.0\ \text{ms}) \approx 0.6 \times 0.0183 = 0.01098\ \text{A} \approx 0.011\ \text{A}$