Solution :
Given $\phi = 20 w/cm^2$
$A= 30\;cm$
$t= 30 \;mins= 30 \times 60\;s$
Total energy falling on the surface in time t is $U= \phi At$
$\qquad= 20 \times 30 \times (30 \times 60)\;J$
Momentum of incident light $= \large\frac{U}{c}$
$\qquad= \large\frac{20 \times 30 \times (30 \times 60)}{3 \times 10^8}$
$\qquad= 36 \times 10^{-4}-0$
$\qquad=36 \times 10^{-4} \; kg m/s$