1. Use the relation between electric field and potential

For one-dimensional variation along the $x$-axis, $E_x = -\frac{dV}{dx}$

Given, $V(x)=\frac{20}{x^2-4}$ where $x$ is in $\mu m$.

2. Differentiate the potential

Write $V(x)=20(x^2-4)^{-1}$

Now differentiate: $\frac{dV}{dx}=20(-1)(x^2-4)^{-2}(2x)$ $\frac{dV}{dx}=\frac{-40x}{(x^2-4)^2}$

Hence, $E_x=-\frac{dV}{dx}=\frac{40x}{(x^2-4)^2}$

3. Evaluate at $x=4\,\mu m$

Substitute $x=4$: $E_x=\frac{40\cdot 4}{(4^2-4)^2}$ $E_x=\frac{160}{(16-4)^2}$ $E_x=\frac{160}{12^2}$ $E_x=\frac{160}{144}$ $E_x=\frac{10}{9}\,\text{volt}/\mu m$

4. Determine the direction

Since $E_x$ is positive, $E_x>0$ so the field is along the $+x$ direction.

5. Match with the options

Thus the electric field at $x=4\,\mu m$ is $\boxed{\frac{10}{9}\,\text{volt}/\mu m \text{ in the } +x \text{ direction}}$

So, Option A is correct.