$=600 \times\left(10^3\right)^2 \,m ^2$

$=6 \times 10^8 \,m ^2$

Average rainfall, $h=2.4 \,m$

Volume of water received by rain, $V$

$=A \times h=6 \times 10^8 \times 2.4 \,m ^3$

Water conserved $=10 \%$ of volume received by rain

$=6 \times 10^8 \times \frac{10}{100} \times 2.4 \,m ^3=1.44 \times 10^8 \,m ^3$

$=1.4 \times 10^8 \times 10^3 L =1.4 \times 10^{11} L$

Percentage of total water consumption received by rain is

$=\frac{1.4 \times 10^{11} \times 100}{1.4 \times 10^{12}}=10 \%$