An isotropic point source of light is suspended h metre vertically...

According to question, the situation is shown in the figure below $E_1=\frac{I}{(L O)^2}=\frac{I}{h^2}...(i)$ The illumination at the edge $A$ is given by $E_2=\frac{I \cos \theta}{(L A)^2}=\frac{I \cos \theta}{\left(h^2+r^2\right)}$...(ii)$From figure,$\cos \theta=\frac{h}{\sqrt{\left(h^2+r^2\right)}} \therefore E_2=\frac{I h}{\left(h^2+r^2\right)^{3 / 2}}$Dividing Eq. (i) by Eq. (ii), we get$$\begin{aligned}\frac{E_1}{E_2} & =\frac{I / h^2}{I h /\left(h^2+r^2\right)^{3 / 2}}=\frac{\left(h^2+r^2\right)^{3 / 2}}{h^3} \\ & =\left(\frac{h^2+r^2}{h^2}\right)^{3 / 2}=\left(1+\frac{r^2}{h^2}\right)^{3 / 2}\end{aligned}$$