b
Total volume of rain drops, recrived \(100\,cm\) in a year by area \(1\,m^2\)

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

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,

\(E = \frac{1}{2}m{v^2}\)

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

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