The weight of a body at the surface of earth is 18 N. The weight of the body at an altitude of 3200 km above the earth's surface is (given, radius of earth $$R_e = 6400$$ km)

We need to find the weight of a body at an altitude of 3200 km, given that its weight at the earth's surface is 18 N.

The acceleration due to gravity at a height $$h$$ above the earth's surface is: $$g' = \frac{g}{\left(1 + \frac{h}{R_e}\right)^2}$$ where $$g$$ is the surface gravity and $$R_e$$ is the radius of the earth.

At $$h = 3200$$ km and $$R_e = 6400$$ km, $$\frac{h}{R_e} = \frac{3200}{6400} = \frac{1}{2}$$, so $$g' = \frac{g}{\left(1 + \frac{1}{2}\right)^2} = \frac{g}{\left(\frac{3}{2}\right)^2} = \frac{g}{\frac{9}{4}} = \frac{4g}{9}$$.

Since weight is proportional to $$g$$, $$W' = W \times \frac{g'}{g} = 18 \times \frac{4}{9} = 8 \text{ N}$$.

The correct answer is Option (4): 8 N.