1. Force due to the original full sphere

Since the point mass $m$ is outside the uniform solid sphere ($2R > R$), the sphere behaves like a point mass $M$ placed at its centre $O$.

So,

2. Mass of the removed spherical part

The big sphere is uniform, so density is constant.

If radius of removed sphere is $R/3$, then its mass is proportional to volume:

3. Location of the removed cavity

From the figure (standard cavity-inside-sphere arrangement), the small removed sphere of radius $R/3$ is tangent to the outer surface along the line joining $O$ and $m$.

Hence the centre of the removed sphere is at distance

from $O$, toward the particle $m$.

Since $m$ is at distance $2R$ from $O$, its distance from the centre $O'$ of the removed spherical part is

4. Force due to the removed part

Again, $m$ lies outside the removed spherical part, so that part acts like a point mass $M/27$ at $O'$.

Thus,

Now,

so

5. Force due to the remaining part

The removed mass was on the same side as the particle, so its gravitational pull was in the same direction as $F_1$. Therefore,

Substitute:

6. Find the ratio $F_1 : F_2$

Cancel common factors:

Multiply both by $48$:

7. Correct option

So the correct option is B.