Let $d$ be range of TV transmission.

From above figure, 

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

 $\Rightarrow d^2=h^2+R^2+2RH-R^2$

$\Rightarrow d^2=h^2+2Rh$

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

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

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

Dividing equation $2$ by $1$, we have

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

$\Rightarrow 9=\frac{h^{\text{'}}}{h}$

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

If $h=100 \mathrm{m}$, then tower of height $900 \mathrm{m}$ is required.