We are given the acceleration \( a = \frac{5}{4} \, \text{ms}^{-2} \) and the body starts from rest. We need to find the distance travelled by the body in the third second.
The formula for the distance travelled in the \( n^{\text{th}} \) second is given by:
\[
S_n = u + \frac{a}{2} (2n - 1),
\]
where:
- \( u \) is the initial velocity (which is 0 since the body starts from rest),
- \( a \) is the acceleration,
- \( n \) is the time in seconds.
We need to find the distance in the third second, so substitute \( n = 3 \) into the equation. The distance travelled in the third second is:
\[
S_3 = u + \frac{a}{2} \left( 2 \times 3 - 1 \right).
\]
Since \( u = 0 \) and \( a = \frac{5}{4} \, \text{ms}^{-2} \), we get:
\[
S_3 = \frac{\frac{5}{4}}{2} \times (6 - 1) = \frac{5}{8} \times 5 = \frac{25}{8} \, \text{m}.
\]
Thus, the distance travelled by the body in the third second is \( \frac{25}{8} \, \text{m} \).
Therefore, the correct answer is option (2).