7GM^2/576R^2.                                                                                                                                                                 .....................................................

Hello Suban Abdul

Given  r = radius of the solid sphere

m = mass of the solid sphere

Taken  d =density of the solid

m' = mass of the removed sphere

M = mass of the remaining body

a = distance between center of cavity and gravitational point of the remaining body.

b = distance between gravitational point of the remaining body & the particle

In first, we have to find values for m' & M.

m = d * (4/3)πr³ = 4dπr³/3

m' = d * (4/3)π(r/2)³ = dπr³/6 ----------- (1)

M = m-m'

=  4dπr³/3 - dπr³/6 = 7dπr³/6 = 7m/8 ------- (2)

Then we can find value of 'a' by taking torque respect to center of gravity of initial sphere,

Mg * (a-r/2) - m'g * (r/2) = 0

M(a-r/2) = m'r/2

(7dπr³/6) * (a-r/2) = (dπr³/6)*(r /2)         [by (1) & (2)]

7 * (a-r/2) = r /2

a = 4r/7 = 0.57r

Then we can write equation for 'b' as below,

b = a + 2.5r

b = 0.57r + 2.5r

b = 3.07r

From the gravitational force formula,

gravitational force = GMm/b² (G =Newtonian constant of gravitation)

= [G * (7m/8 ) *m ] / (3.07r²)

= (6.674 08 x 10-11 * 7m²)/24.56r²

= 1.0902 x 10-11m²/r²

Hope it helps you.