Aman stand btn 2 clift and fires agun ,he hears the first echo after the 2 second and the second echo after 3/2 second.caculate the distance between the 2clift

The speed of sound in dry air at $20^\circ\text{C}$ is approximately $v = 343\ \text{m/s}$.

The first echo is heard after $t_1 = 2\ \text{s}$. The sound travels to the cliff and back, covering twice the distance to the cliff.
Let $d_1$ be the distance to the first cliff.
$2d_1 = v \times t_1$
$d_1 = \frac{v \times t_1}{2}$
$d_1 = \frac{343\ \text{m/s} \times 2\ \text{s}}{2}$
$d_1 = 343\ \text{m}$

The second echo is heard after $t_2 = \frac{3}{2}\ \text{s} = 1.5\ \text{s}$.
Let $d_2$ be the distance to the second cliff.
$2d_2 = v \times t_2$
$d_2 = \frac{v \times t_2}{2}$
$d_2 = \frac{343\ \text{m/s} \times 1.5\ \text{s}}{2}$
$d_2 = \frac{514.5\ \text{m}}{2}$
$d_2 = 257.25\ \text{m}$

The total distance between the two cliffs, $D$, is the sum of the distances from the man to each cliff.
$D = d_1 + d_2$
$D = 343\ \text{m} + 257.25\ \text{m}$
$D = 600.25\ \text{m}$

Rounding to three significant figures (based on the speed of sound), the distance is $600.\ \text{m}$.