The approximate height from the surface of earth at which the weight of the body becomes $$\frac{1}{3}$$ of its weight on the surface of earth is :
[Radius of earth $$R = 6400$$ km and $$\sqrt{3} = 1.732$$]

We need to find the height $$h$$ from the surface of the Earth where the weight becomes $$\frac{1}{3}$$ of its weight on the surface.

Write the formula for gravitational acceleration at height $$h$$: $$ g' = \frac{g}{\left(1 + \frac{h}{R}\right)^2} $$

Set up the equation: for weight to become $$\frac{1}{3}$$:

$$ \frac{g}{3} = \frac{g}{\left(1 + \frac{h}{R}\right)^2} $$

$$ \left(1 + \frac{h}{R}\right)^2 = 3 $$

$$ 1 + \frac{h}{R} = \sqrt{3} $$

Solve for $$h$$: $$ \frac{h}{R} = \sqrt{3} - 1 $$

$$ h = R(\sqrt{3} - 1) = 6400 \times (1.732 - 1) $$

$$ h = 6400 \times 0.732 = 4684.8 \approx 4685 \text{ km} $$

Therefore, the correct answer is Option B.