Answer
$d =15 \times 10^{5} \mathrm{m}$
Work Step by Step
The potential difference between the two plates is due to the battery connected to the plates and the distance between the two plates could be calculated by
\begin{gather}
V =E d \\
d=\frac{V}{E}\\
d =\frac{1.5 \mathrm{V}}{1 \times 10^{-6} \mathrm{V} / \mathrm{m}}\\
\boxed{d =15 \times 10^{5} \mathrm{m}}
\end{gather}
