Distance covered in the last one second i.e., distance covered in the nth second if the ball takes n seconds to hit the ground is 
$x_1=S_{n^{th}}=0+\frac{g}{2}(2n-1)$    ….(1)
Distance covered in the first three seconds is 
$X_2=\frac{1}{2}g{\left(3\right)}^2=\frac{9g}{2}$     ….(2)
Since  $X_1=X_2$
$\Rightarrow \frac{g}{2}(2n-1)=\frac{9g}{2}$
$\Rightarrow 2n-1=9$
$\Rightarrow n=5s$
So, total time taken by the ball to hit the ground is 5 s. So,
$h=\frac{1}{2}g{t}^2$
$\Rightarrow h=\frac{1}{2}\left(10\right){\left(5\right)}^2=5\left(25\right)=125m$
$\Rightarrow h=125\text{\hspace{0.17em}m}$