Solution:

We will use the formula for the distance covered in the \(n\)-th second:

\[
s_n = u + \frac{a}{2}(2n - 1)
\]

Given:
- Distance in the 5th second: \( s_5 = 65 \, \text{m} \)
- Distance in the 9th second: \( s_9 = 105 \, \text{m} \)

Using the formula for both:
1. \( s_5 = u + \frac{a}{2}(9) = 65 \)
\[
u + \frac{9a}{2} = 65 \tag{1}
\]

2. \( s_9 = u + \frac{a}{2}(17) = 105 \)
\[
u + \frac{17a}{2} = 105 \tag{2}
\]

Now, subtract equation (1) from equation (2):
\[
\left( u + \frac{17a}{2} \right) - \left( u + \frac{9a}{2} \right) = 105 - 65
\]
\[
\frac{8a}{2} = 40 \implies 4a = 40 \implies a = 10 \, \text{m/s}^2
\]

Substitute \( a = 10 \) into equation (1):
\[
u + \frac{9 \times 10}{2} = 65 \implies u + 45 = 65 \implies u = 20 \, \text{m/s}
\]

Now, to find the total distance covered in 20 seconds:
\[
S = ut + \frac{1}{2} a t^2 = 20 \times 20 + \frac{1}{2} \times 10 \times 20^2
\]
\[
S = 400 + 0.5 \times 10 \times 400 = 400 + 2000 = 2400 \, \text{m}
\]

Thus, the body will cover 2400 meters in 20 seconds.