At t = 0 , the plates of a capacitor of capacitance 10 ext ,...

The discharging of a capacitor is given by the equation:

$q(t) = q_0 e^{-t / \tau}$

where $\tau = RC$ is the time constant of the circuit.

Given values are $C = 10 \mu\text{F}$, $R = 10 \Omega$, $q_0 = 60 \mu\text{C}$, and $t = 1.0 \text{ ms} = 10^{-3} \text{ s}$.

Calculating the time constant $\tau$:

$\tau = RC = 10 \times 10 \times 10^{-6} = 10^{-4} \text{ s}$

Substituting the values into the discharging equation:

$q(t) = 60 \times e^{-10^{-3} / 10^{-4}} \mu\text{C}$

$q(t) = 60 \times e^{-10} \mu\text{C}$

Using the approximate value $e^{10} \approx 22026$:

$q(t) = \dfrac{60}{22026} \mu\text{C} \approx 0.00272 \mu\text{C}$

Rounding to the nearest given option, the charge is approximately $0.003 \mu\text{C}$.