(a)
Hint: \(Capacitance,~C=\frac{\epsilon _{0}A}{d}~and~q=CV\)
Step 1: Find capacitance.
The radius of each circular plate (r)$=12 \mathrm{cm}=\frac{12}{100} \mathrm{m}=0.12 \mathrm{m}$
The distance between the plates (d)$=5 \mathrm{cm}=\frac{5}{100} \mathrm{m}=0.05 \mathrm{m}$

Capacitance is given, $C=\frac{{\in }_{\circ }A}{d}$ where A =${\mathrm{\pi r}}^2$

$\mathrm{C}=\frac{{\in }_0{\mathrm{\pi r}}^2}{\mathrm{d}}=\frac{8.85\times {10}^{-12}\times \mathrm{\pi }\times {\left(0.12\right)}^2}{0.05}=8.0032\times {10}^{-12}\mathrm{}F=8.0032pF$

Step 2: Find the rate of change of potential difference between the plates.
Charge on each plate, q = CV
Differentiating on both sides:
$$\(\frac{dq}{dt}=C\frac{dV}{dt}\)

$dVdt=IC=0.158.0032\times 10-12=1.87\times 1010V/s$

(b)
Hint: The displacement current across the plates is equal to the conduction current.
Step: Find displacement current.
The displacement current is 0.15 A.

(c) 
Hint: Recall Kirchhoff’s first law
Step: Take the sum of conduction and displacement current.
Yes, Kirchhoff’s first law (junction law) is valid at each plate of capacitor. We take into consideration both the conduction and displacement current.