The question can be divided into two parts, where in the first part, the time taken by the pebble to reach the bottom is calculated and in the second part, the time taken by the sound wave to reach the top is calculated.
Given: 
Initial velocity, $u=0m{s}^{-1}$
Depth of the well, $s=100m$
Acceleration, $a=10m{s}^{-2}$
Speed of sound, $v=343m{s}^{-1}$
Let $t_1$ be the time taken by the pebble to hit the bottom.
From second equation of motion,
$\text{s}=u{t}_1+\frac{1}{2}a{t}_1^2$
$100=(0\times t_1)+(\frac{1}{2}\times (10)\times t_1^2)$
$100=5t_1^2$
$\Rightarrow t_1=\sqrt{20}s$
We know, $time=\frac{distance}{speed}$
Let $t_2$ be the time taken for the sound wave to reach from the bottom to the top of the well.
$t_2=\frac{s}{v}=\frac{100}{343}\approx 0.3s$
Total time,
$t=t_1+t_2$
  $=\sqrt{20}+0.3s$.