$ = 1{m^2} \times \frac{{100}}{{100}}m = 1\,{m^3}$

As we know, density of water,

$d = {10^3}\,kg/{m^3}$

Therefore, mass of this volume of water,

$M = d \times v = {10^3} \times 1 = {10^3}\,kg$

Average terminal velocity of rain drop 

$v = 9\,m/s\,\left( {given} \right)$

Therefore, energy transferred by rain,

$E = \frac{1}{2}m{v^2}$

$ = \frac{1}{2} \times {10^3} \times {\left( 9 \right)^2}$

$ = \frac{1}{2} \times {10^3} \times 81 = 4.05 \times {10^4}J$