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**Hint**Acceleration due to gravity is defined as the acceleration which is gained by an object due to the gravitational force. The SI unit is given as a meter per second square. Based on this concept we can solve this question.

**Complete step by step answer**

Suppose, the mass of the core of the planet is m and that of the outer shell is m.

So, the field on the surface of the core is $\dfrac{Gm}{{{R}^{2}}}$.

And, on the surface of the shell it will be $\dfrac{G(m+{{m}^{/}})}{{{(2R)}^{2}}}$.

So, we can find that:

$\dfrac{Gm}{{{R}^{2}}}=\dfrac{G(m+{{m}^{/}})}{{{(2R)}^{2}}}$

So, we can say that:

$4m=m+{{m}^{/}}$

$\Rightarrow {{m}^{/}}=3m$

Let us consider that $m=\dfrac{4}{3}\pi {{R}^{3}}{{p}_{1}}$

So, we can evaluate that:

$3m=3\left( \dfrac{4}{3}\pi {{R}^{3}}{{p}_{1}} \right)={{m}^{/}}=\dfrac{4}{3}\pi 7{{R}^{3}}{{p}_{2}}$

So, the value we get that:

${{p}_{1}}=\dfrac{7}{3}{{p}_{2}}$

**So, the ratio of ${{p}_{1}}/{{p}_{2}}$is given as $7/3$.**

**Note**We should know that the gravitational force depends on two factors : the mass and the distance. If the mass of one object is doubled then force of gravity between the objects also gets doubled. Increasing the distance, results in the lowering of the gravitational force.

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