(N/A) Radius of each circular plate,$r = 12 \; cm = 0.12 \; m$
Distance between the plates,$d = 5 \; cm = 0.05 \; m$
Charging current,$I = 0.15 \; A$
Permittivity of free space,$\varepsilon_{0} = 8.85 \times 10^{-12} \; C^{2} \; N^{-1} \; m^{-2}$
$(a)$ Capacitance $C$ is given by $C = \frac{\varepsilon_{0} A}{d}$,where $A = \pi r^{2}$.
$C = \frac{8.85 \times 10^{-12} \times \pi \times (0.12)^{2}}{0.05} \approx 8.0032 \times 10^{-12} \; F = 80.032 \; pF$.
Since $q = CV$,differentiating with respect to time gives $\frac{dq}{dt} = C \frac{dV}{dt}$.
Given $\frac{dq}{dt} = I$,we have $\frac{dV}{dt} = \frac{I}{C} = \frac{0.15}{80.032 \times 10^{-12}} \approx 1.87 \times 10^{9} \; V/s$.
$(b)$ The displacement current $i_{d}$ between the plates is equal to the conduction current $I$ in the connecting wires. Thus,$i_{d} = 0.15 \; A$.
$(c)$ Yes,Kirchhoff's first rule is valid at each plate of the capacitor if we consider the total current as the sum of conduction current and displacement current.