Gravitational field intensity of a carved out spherical body

Relevant wiki: Gravitation

Let the initial mass of the sphere is $M_{s}.$

So the mass of the spherical cavity $(m)$ can be found by comparing the densities,

$\frac{3m}{4πR^{3}} = \frac{3M_{s}}{32πR^{3}}.$

$m = \frac{M_{s}}{8}.$

Gravitation field due to a sphere at a distance $r >R$ is $E = \dfrac{GM}{r^2}.$
Thus, the total gravitational field intensity due to whole sphere at A is,

$E = \frac{GM_{s}}{36R^{2}}.$

Gravitational intensity at A due the sphere which has been removed,

$E_{c} = \frac{GM_{s}}{200R^{2}}.$

So, the gravitational intensity at A due to the remaining part of the sphere is,

$\begin{aligned} E_{r} &=& E - E_{c}. \\ E_{r} &=& \frac{GM_{s}}{36R^{2}} - \frac{GM_{s}}{200R^{2}}. \\ E_{r} &=& \frac{41GM_{s}}{1800R^{2}}. \\ \end{aligned}$

So the remaining mass is,

$\begin{aligned} M &=& M_{s} - \frac{M_{s}}{8}. \\ M &=& \frac{7M_{s}}{8}. \\ M_{s} &=& \frac{8M}{7}. \\ E_{r} &=& \boxed{\frac{41GM}{1575R^{2}}}. \\ \end{aligned}$