Answer
$i_{d}=2.87×10^{16}A$
Work Step by Step
The displacement current is
$
\begin{aligned}
i_d & =\varepsilon_0 \frac{d \Phi_E}{d t}=\varepsilon_0 A \frac{d E}{d t}=\varepsilon_0 A \frac{d}{d t}\left(\frac{\rho i}{A}\right)=\varepsilon_0 \rho \frac{d i}{d t}=\left(8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)\left(1.62 \times 10^{-8} \Omega\right)(2000 \mathrm{~A} / \mathrm{s}) \\
& =2.87 \times 10^{-16} \mathrm{~A} .
\end{aligned}
$
