The gravitational field, due to the 'left over part' of a uniform...

Let mass of smaller sphere (which has to be removed) is $\mathrm{m}$ Radius $=\frac{\mathrm{R}}{2}($ from figure $)$ $$\begin{aligned}& \frac{\mathrm{M}}{\frac{4}{3} \pi \mathrm{R}^3}=\frac{\mathrm{m}}{\frac{4}{3} \pi\left(\frac{\mathrm{R}}{2}\right)^3}\\ & \Rightarrow \mathrm{m}=\frac{\mathrm{M}}{8} \end{aligned}$$Mass of the left over part of the sphere \mathrm{M}^{\prime}=$\mathrm{M}-\frac{\mathrm{M}}{8}=\frac{7}{8} \mathrm{M}$Therefore gravitational field due to the left over part of the sphere =$\frac{\mathrm{GM}^{\prime}}{\mathrm{x}^2}=\frac{7}{8} \frac{\mathrm{GM}}{\mathrm{x}^2}$