a
Here, \(V(x)=\frac{20}{x^{2}-4}\) \(volt\)

We know that \(E=-\frac{d V}{d x}=-\frac{d}{d x}\left(\frac{20}{x^{2}-4}\right)\)

or, \(E=+\frac{40 x}{\left(x^{2}-4\right)^{2}}\)

At \(x=4\, \mu \mathrm{m}\)

\(E=+\frac{40 \times 4}{\left(4^{2}-4\right)^{2}}=+\frac{160}{144}=+\frac{10}{9}\) \(volt / \mu \mathrm{m}\)

Positive sign indicates that \(\vec{E}\) is in \(+ ve\, x-\) direction.