What is the height from the surface of earth,where acceleration due to gravity will be $\frac{1}{4}$ of that of the earth (in $km$)? $(R_E = 6400 \ km)$

Given below are two statements: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A:$ $A$ pendulum clock when taken to Mount Everest becomes fast.
Reason $R:$ The value of $g$ (acceleration due to gravity) is less at Mount Everest than its value on the surface of earth.
In the light of the above statements,choose the most appropriate answer from the options given below.

$A$ pendulum is oscillating with frequency $n$ on the surface of the Earth. If it is taken to a depth $d = \frac{R}{4}$ below the surface of the Earth,what is the new frequency of oscillation? ($R =$ radius of the Earth)

Imagine a new planet having the same density as that of the earth, but it is two times bigger than the earth in size (radius). If the acceleration due to gravity on the surface of the new planet is $g'$ and that on the surface of the earth is $g$, then:

How much is the radius of the Earth at the equator greater than the radius at the poles?

Two spherical planets $A$ and $B$ have the same mass,but their densities are in a ratio $8:1$. For these planets,the ratio of acceleration due to gravity at the surface of $A$ to its value at the surface of $B$ is: