The required height of a TV tower which can cover the population of...

Maximum range of TV transmission $d = \sqrt{2Rh}$

$d = \sqrt{2 \times 6400 \times h \times {10}^{- 3}}$ (here, $h$ in $m$)

Maximum area of TV transmission

Area $= \pi{d}^{2}$

$= \left(\pi \times 2 \times 6400 \times h \times {10}^{- 3}\right){km}^{2}$

Now, $6 . 03 \times 100000 = 100 \times \pi \times 2 \times 6400 \times {10}^{- 3}h$

$h = \dfrac{6 . 03 \times {10}^{5}}{10 \times \pi \times 128}$

$h = 150m$