At a particular station, the TV transmission tower has a height of...

Let $d$ be range of TV transmission.

From above figure,

${d}^{2} = {\left(h + R\right)}^{2} - {R}^{2}$

$⇒ {d}^{2} = {h}^{2} + {R}^{2} + 2RH - {R}^{2}$

$⇒ {d}^{2} = {h}^{2} + 2Rh$

As $R ≫ ≫ h$ then, $d \approx \sqrt{2Rh} . . . \left(1\right)$

Now, if coverage is to be increased $3$ times

$3d = \sqrt{2R{h}^{'}} . . . \left(2\right)$

Dividing equation $2$ by $1$, we have

$\dfrac{3d}{d} = \sqrt{\dfrac{2R{h}^{'}}{2Rh}}$

$⇒ 9 = \dfrac{{h}^{'}}{h}$

$9h = {h}^{'}$

If $h = 100m$, then tower of height $900m$ is required.