Answer
$1.7\times10^{15}\frac{V/m}{s}$
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.
$$I_D=\epsilon_o\frac{\Delta \Phi_E}{\Delta t}$$
$$ I_D =\epsilon_o A\frac{\Delta E}{\Delta t}$$
The current in the wires leading to the plates is equal to the displacement current.
Solve for the rate of change of the electric field.
$$ \frac{\Delta E}{\Delta t}=\frac{I_D }{\epsilon_o A}$$
$$=\frac{3.8A}{(8.85\times10^{-12}C^2/(N \cdot m^2))(0.0160m)^2}$$
$$=1.7\times10^{15}\frac{V/m}{s}$$