The total energy falling on the surface is $ U = (18\: W/cm^2 ) \times ( 20\: cm^2) \times (30 \times 60) = 6.48 \times 10^5J$
Therefore, the total momentum delivered (for complete absorption) is
$ p = \large\frac{U}{c}$$= \large\frac{(6.48 \times 10^5J) }{3 \times 10^8 m/s)}$ $ =m/s) = 2.16 \times 10^{-3}\: kgm/s$
The average force exerted on the surface is
$ F = \large\frac{p}{t}$$ \large\frac{2.16 \times 10^{-3}}{0.18 \times 10^4}$$ = 1.2 \times 10^{-6}N$
Ans : (d)