Answer
$=8.59 \times 10^5 \mathrm{~V} \cdot \mathrm{m} / \mathrm{s} .
$
Work Step by Step
As $i_d=a_0\left(d \Phi_E / d t\right)$, we have
$$
\left\lceil\frac{d \Phi_E}{d t}\right\}_{\max }=\frac{i_{d \max }}{\varepsilon_0}\\=\frac{7.60 \times 10^{-6} \mathrm{~A}}{8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}}\\=8.59 \times 10^5 \mathrm{~V} \cdot \mathrm{m} / \mathrm{s} .
$$
