Answer
$\frac{4λaz}{4πϵ_{^{\circ}} (z^2 + \frac{a^2}{4} ) \sqrt (z^2+\frac{a^2}{2})}$
Work Step by Step
From Ex. 2.2 $\overset{\rightharpoonup}{E}$ =$\frac{2λL}{4πϵ_{^{\circ}} x\sqrt (x^2+L^2)}$ is defined for rod of length $2L$
A square loop can be considered as its made from 4 charged rod of lenght a each.
So, $L=a/2$
Consider a single straight section of square loop of length a
From the point it is at a perpendicular distance of $\sqrt (z^2 + (a/2)^2)$
Therefore substituting $L=a/2$ and $x= \sqrt (z^2 + (a/2)^2)$ in the given formula
We get,
$\overset{\rightharpoonup}{e}=\frac{λL}{4πϵ_{^{\circ}} \sqrt (z^2 + \frac{a^2}{4} ) \sqrt (z^2+\frac{a^2}{2})}$
Now the resultant Electric fied due to all four rod will have component only in $\hat z$ direction as a result of vector addition.
The component in $\hat z$ direction : $\overset{\rightharpoonup}{E}=4 (\overset{\rightharpoonup}{e} cos\theta $)
$cos \theta = \frac{z}{\sqrt (z^2 + \frac{a^2}{4})}$
Hence the resultant Electric field due to a square loop is
$\overset{\rightharpoonup}{E}=\frac{4λaz}{4πϵ_{^{\circ}} (z^2 + \frac{a^2}{4} ) \sqrt (z^2+\frac{a^2}{2})}$
