Force exerted by the beam on the surface is given by $F=\frac{∆p}{∆t}=\frac{2IA\cos \theta }{c}$

Given here: $I={10}^{-3 }\mathrm{N} {\mathrm{m}}^{-2}, A=20 {\mathrm{cm}}^2$ and $\theta =45°$.

Putting the values, we get

$F=\frac{2\times {10}^{-3}\times 20\times {10}^{-4}\times \cos 45°}{3\times {10}^8}=9.4\times {10}^{-15 }\mathrm{N}$