A

To find the ratio of distances traveled by a freely falling body in different time intervals, we can use the equation of motion under gravity. The distance traveled in a given time under constant acceleration can be expressed as:

$d = \frac{1}{2} g t^2$, where $d$ is the distance, $g$ is the acceleration due to gravity, and $t$ is the time.

For the first second:
$d_1 = \frac{1}{2} g (1^2) = \frac{1}{2} g$
For the second second (from 1 s to 2 s):
$d_2 = \frac{1}{2} g (2^2) - \frac{1}{2} g (1^2) = 2g - \frac{1}{2} g = \frac{3}{2} g$
For the third second (from 2 s to 3 s):
$d_3 = \frac{1}{2} g (3^2) - \frac{1}{2} g (2^2) = \frac{9}{2} g - 2g = \frac{5}{2} g$
For the fourth second (from 3 s to 4 s):
$d_4 = \frac{1}{2} g (4^2) - \frac{1}{2} g (3^2) = 8g - \frac{9}{2} g = \frac{7}{2} g$

The distances from 1st to 4th second are therefore:
d1:d2:d3:d4 = (1:3:5:7)

Consolidating this to the ratio of distances traveled yields the final result. Therefore, the correct answer, based on the given options, is A.