The radius in kilometer to which the present radius of earth ($$R = 6400$$ km) to be compressed so that the escape velocity is increased 10 times is ________.

We need to find the compressed radius $$R'$$ of the Earth such that the escape velocity becomes 10 times its current value, while the mass remains the same.

The escape velocity from the surface of a planet is given by $$v_e = \sqrt{\frac{2GM}{R}}$$, where $$M$$ is the mass and $$R$$ is the radius.

Let the new escape velocity be $$v_e' = 10 v_e$$. Since $$v_e' = \sqrt{\frac{2GM}{R'}}$$ and $$v_e = \sqrt{\frac{2GM}{R}}$$, we get $$\frac{v_e'}{v_e} = \sqrt{\frac{R}{R'}}$$.

Substituting $$v_e' = 10 v_e$$, we have $$10 = \sqrt{\frac{R}{R'}}$$, which gives $$100 = \frac{R}{R'}$$, so $$R' = \frac{R}{100}$$.

With $$R = 6400$$ km, the compressed radius is $$R' = \frac{6400}{100} = 64$$ km.

Therefore, the required compressed radius is $$\boxed{64}$$ km.