Answer
$4.8\times10^{-8}A$
Work Step by Step
On page 627, the displacement current is derived as $I_D=\epsilon_o\frac{\Delta \Phi_E}{\Delta t}$.
The electric field flux is given by EA, where A is the area of the capacitor plates.
$$I_D=\epsilon_o\frac{\Delta \Phi_E}{\Delta t}$$
$$=\epsilon_o A\frac{\Delta E}{\Delta t}$$
$$=(8.85\times10^{-12}C^2/(N \cdot m^2))(0.058m)^2(1.6\times10^6\frac{V/m}{s})$$
$$=4.8\times10^{-8}A$$
