Explanation

(a) $h = 150\ m,\ R = 6.4 \times 10^{6}\ m$
Average population density $= 10^{3}\ km = 10^{3} \times \left( 10^{3} \right)^{- 2}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 10^{- 3}\ m$
Distance up to which the transmission would be view
$d = \sqrt{2hR}$
Total area over which transmission could be
$= \pi d^{2} = 2\pi hR$
Population covered $= 10^{- 3} \times 2\pi hR$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 10^{- 3} \times 2 \times 3.00 \times 150 \times 6.4 \times 10^{- 6}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 60.288\ lakhs$