Answer
$V \approx\dfrac{\pi k_e \sigma R^2}{d}$ for large $d$
Work Step by Step
Consider $\sqrt{d^2+R^2}=\sqrt{d^2(1+\dfrac{R^2}{d^2})}$
or, $=d(1+\dfrac{R^2}{d^2})^{1/2}$
Apply the Binomial series.
$d(1+\dfrac{R^2}{d^2})^{1/2}=d(1+\dfrac{R^2}{2d^2}+...)=d+\dfrac{R^2}{2d}+....$
Consider the given expression for potential.
$V \approx 2\pi k_e \sigma (d+\dfrac{R^2}{2d}+......-d)$
Re-write as
$V \approx 2\pi k_e \sigma (\dfrac{R^2}{2d})$
Hence, $V \approx\dfrac{\pi k_e \sigma R^2}{d}$ for large $d$