Answer
$\frac{d|\vec{E}|}{d t}=-8.8 \times 10^{15} \mathrm{~V} / \mathrm{m} \cdot \mathrm{s} $
Work Step by Step
Using Formula but noting that the capacitor is being discharged, we have
$$
\frac{d|\vec{E}|}{d t}=-\frac{i}{\varepsilon_0 A}\\=-\frac{5.0 \mathrm{~A}}{\left(8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{N} \cdot \mathrm{m}^2\right)(0.0080 \mathrm{~m})^2}\\=-8.8 \times 10^{15} \mathrm{~V} / \mathrm{m} \cdot \mathrm{s} .$$
