If earth has a mass nine times and radius twice to that of a planet P. Then $$\frac{v_{e}}{3} \sqrt{x} \mathrm{~ms}^{-1}$$ will be the minimum velocity required by a rocket to pull out of gravitational force of $$\mathrm{P}$$, where $$v_{e}$$ is escape velocity on earth. The value of $$x$$ is

Given below are two statements:

Statement I: Acceleration due to gravity is different at different places on the surface of earth.

Statement II: Acceleration due to gravity increases as we go down below the earth's surface.

In the light of the above statements, choose the correct answer from the options given below

At a certain depth "d " below surface of earth, value of acceleration due to gravity becomes four times that of its value at a height $$\mathrm{3 R}$$ above earth surface. Where $$\mathrm{R}$$ is Radius of earth (Take $$\mathrm{R}=6400 \mathrm{~km}$$ ). The depth $$\mathrm{d}$$ is equal to