To find the gravitational force on a body at a height equal to one-third the Earth's radius from its surface, we start with the weight of the body on the Earth’s surface, which is 48 N.

The gravitational force at the surface, $ W $, is given by:

$ W = mg $

Where $ g = \frac{GM}{R^2} $.

At a height $ h $ above the Earth's surface, the gravitational force $ g_h $ is:

$ g_h = \frac{GM}{(R+h)^2} $

To find the weight $ W_h $ at this height, the ratio of gravitational forces at height $ h $ and at the surface is:

$ \frac{W_h}{W} = \frac{g_h}{g} = \frac{R^2}{(R+h)^2} $

Given $ h = \frac{R}{3} $, substitute $ h $ into the equation:

$ \frac{W_h}{W} = \frac{R^2}{\left(R+\frac{R}{3}\right)^2} = \frac{R^2}{\left(\frac{4R}{3}\right)^2} = \frac{9}{16} $

Thus, the new weight $ W_h $ is:

$ W_h = \frac{9}{16} \times W = \frac{9}{16} \times 48 \, \text{N} $

$ W_h = 27 \, \text{N} $

Therefore, at this height, the gravitational force experienced by the body is 27 N.