Answer
$\left|i_d\right|=0.71 A$
Work Step by Step
In region $a$ of the graph,
$
\left|i_d\right|=\varepsilon_0\left|\frac{d \Phi_E}{d t}\right|\\=\varepsilon_0 A\left|\frac{d E}{d t}\right|\\=\left(8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right)\left(1.6 \mathrm{~m}^2\right)\left|\frac{4.5 \times 10^5 \mathrm{~N} / \mathrm{C}-6.0 \times 10^5 \mathrm{~N} / \mathrm{C}}{4.0 \times 10^{-6} \mathrm{~s}}\right|\\=0.71 \mathrm{~A}
$
