\({C}=\frac{\epsilon_{0} {A}}{{d}}={k} \frac{\epsilon_{0} \times 1}{2 \times 10^{-3}} {F}\)

\(\therefore {X}_{{c}}=\frac{1}{\omega {C}}=\frac{2 \times 10^{-3}}{2 \times 50 \pi {\epsilon}_{0}}=\frac{2 \times 10^{-3}}{25 \times 4 \pi \epsilon_{0}} \Omega\)

\(\therefore {X}_{{c}}=\frac{2 \times 10^{-3}}{25} \times 9 \times 10^{9}=\frac{18}{25} \times 10^{6} \Omega\)

\(\therefore {i}_{0}=\frac{{V}_{0}}{{X}_{{c}}}=\frac{20 \times 25}{18} \times 10^{-6} {A}=27.47 \mu {A}\)

The value of amplitude of displacement current will be same as value of amplitude of conventional current.