The distance covered in first 3 second
$s=\large\frac{1}{2} $$gt^2$
$\quad=\large\frac{1}{2} $$ \times 10 \times 3^2$
$ \quad=45$
If the ball takes 'n' second to fall to ground . The distance covered in $n^{th}$ second
$s_n=u+\large\frac{g}{2}$$(2n-1)$
$\quad=0+\large\frac{10}{2}$$(2n-1)$
$\quad=10n-5$
Therefore $45=10n-5$
$n=5$
Therefore $ h=\large\frac{1}{2} $$gt^2$
$\qquad=\large\frac{1}{2} $$\times 10 \times 25$
$\qquad=125 \;m$
Hence a is the correct answer.