Here, we have to calculate the average force calculated at the surface.

Given

I = $25\times {10}^4$

A = $15\times {10}^{-4}$

We know that,

$F=\frac{2IA}{C}=\frac{2\left(25\times {10}^4\right)\left(15\times {10}^{-4}\right)}{3\times {10}^8}=2.5\times {10}^{-6}\mathrm{N}$

Hence, force will be $2.5\times {10}^{-6}\mathrm{N}$

So Option (2) will be correct.