According to the question, a parallel plate capacitor is shown in the figure below,


Given, radius \(=2 \mathrm{~cm}\), distance between plates,
\(d=0.1 \mathrm{~mm} \text { and } \frac{d V}{d t}=5 \times 10^6 \mathrm{Vs}^{-1}\)
Displacement current in the capacitor,
\(\begin{aligned}
I_d=\varepsilon_0 \frac{d \phi}{d t} & =\varepsilon_0 \frac{d E}{d t} A \\
I_d & =\varepsilon_0 A \frac{d E}{d t} \\
\Rightarrow \quad I_d & =\varepsilon_0 \frac{A}{d} \frac{d V}{d t} \quad\left(\because E=\frac{V}{d}\right)
\end{aligned}\)
Putting the given values, we get
\(\begin{aligned}
& \Rightarrow \quad I_d=\frac{8.85 \times 10^{-12} \times \pi \times\left(4 \times 10^{-4}\right) \times 5 \times 10^6}{1 \times 10^{-4}} \mathrm{~A} \\
& \Rightarrow \quad I_d=556.28 \times 10^{-6} \mathrm{~A}=0.556 \mathrm{~mA}
\end{aligned}\)
Hence, the correct option is (c).