A uniform ring of mass m is lying at a distance $\sqrt{3}$a form the centre of a sphere of mass M just over the sphere (where a is the radius of the ring as well as that of the sphere). Then magnitude of gravitational force between them is 

For a small mass dm, distance from the center of sphere is 2a. force along ‘dm’  $\frac{\mathrm{G},\mathrm{M},\mathrm{dm}}{4{\mathrm{a}}^2}=\mathrm{F}$
Component of force F sin   get cancelled net force 2Fcos $\mathrm{\theta }$ . 
For the entire ring- $\frac{\mathrm{GM}}{4{\mathrm{a}}^2}\frac{\mathrm{m}}{2}\times 2\mathrm{cos\theta }$

$\frac{\mathrm{GMm}}{4{\mathrm{a}}^2}\frac{\sqrt{3}\mathrm{a}}{2\mathrm{a}}=\frac{\sqrt{3}\mathrm{GMm}}{8{\mathrm{a}}^2}$

Alternative solution:-

field intensity due the ring at the centre of the sphere is

 $\mathrm{g}=\frac{\mathrm{GMx}}{{({\mathrm{R}}^2+{\mathrm{x}}^2)}^{{\displaystyle \frac{3}{2}}}}=\frac{\mathrm{GMa}}{{({\mathrm{a}}^2+3{\mathrm{a}}^2)}^{{\displaystyle \frac{3}{2}}}}=\frac{\sqrt{3}\mathrm{GM}}{8{\mathrm{a}}^2}$

$\mathrm{F}=\mathrm{mg}=\frac{\sqrt{3}\mathrm{GMm}}{8{\mathrm{a}}^2}$