A man stands in front of a large wall at a distance of 50.0 m and...

Let the distance of the man from the wall be $d = 50.0 \text{ m}$.

The sound of the clap travels to the wall and reflects back to the man. The total distance traveled by the sound to hear the echo is $2d = 100 \text{ m}$.

Let the velocity of sound be $v$. The time taken for the echo to reach the man is:
$t = \dfrac{2d}{v}$

The man progressively reduces the time interval between claps until the echo of a clap merges exactly with the subsequent clap. This means the time taken for the echo to return is exactly equal to the time interval between two successive claps.

The man claps $10$ times during every $3$ seconds, which implies the frequency of clapping is $f = \dfrac{10}{3} \text{ Hz}$.

The time interval between successive claps is:
$T = \dfrac{1}{f} = \dfrac{3}{10} = 0.3 \text{ s}$

Equating the echo time to the clap interval:
$\dfrac{2d}{v} = T$

$\dfrac{100}{v} = 0.3$

$v = \dfrac{100}{0.3} = 333.33 \text{ m/s}$

Rounding to the nearest integer, the velocity of sound is $333 \text{ m/s}$.