Solution:

The time taken for an object to fall freely from a height \( h \) is given by the equation:

\[
h = \frac{1}{2} g t^2,
\]

where \( g \) is the acceleration due to gravity.

1. For Earth:
\[
h = \frac{1}{2} g_E t^2 ; t = \sqrt{\frac{2h}{g_E}}.
\]

2. For the Moon, where \( g_M \approx \frac{1}{6} g_E \):
\[
h = \frac{1}{2} g_M t_m^2 ; t_m = \sqrt{\frac{2h}{g_M}} = \sqrt{\frac{2h}{\frac{1}{6} g_E}} = \sqrt{12 \cdot \frac{2h}{g_E}}.
\]

This can be simplified using the time \( t \) from Earth:

\[
t_m = \sqrt{6} \cdot t = \sqrt{6} \cdot t.
\]

Thus, the time taken by the same body to reach the Moon's surface is approximately \( \sqrt{6} t \).