A difficult question.

Let the well be x m deep.
The sound takes (x/343)s to come out from the well bottom.
So the stone takes (3-x/343)s to fall into the well bottom.
s = ut + 1/2 at²
x = 0 + 1/2 9.8(3-x/343)²
2x/9.8 = 9-6x/343+x²/343²
24010x = 1058841-2058x+x²
x²-26068x+1058841 = 0
x = 26067.3(rejected) or x = 40.7
Therefore the well is 40.7m deep.

When the stone is falling,

s = ut + (1/2)at^2

since u = 0 & a = 9.8

so
s = 4.9t^2 (equation 1)

When the stone hits the ground, the sound produced travels upwards.

v = d / (3 – t)

since speed of sound is 343,

so
343 = s / (3 – t)
s = 343 x (3 – t) (equation 2)

put (2) into (1), we have
343 x (3 – t) = 4.9t^2
4.9t^2 + 343t – 1029 = 0
t^2 +70t – 210 = 0
t = 2.88 s

so
put it into equation 2,
s = 343 x (3 – 2.88)
s = 40.7 m