Moment of Inertia (M.I.) of four bodies having same mass M and...

$∵ 2\left({I}_{2} + {I}_{3}\right) + {I}_{4} = x{I}_{1}$

Now

For solid sphere about its diameter ${I}_{1} = \dfrac{2}{5}M{R}^{2}$,

For solid cylinder about its axis ${I}_{2} = \dfrac{1}{2}M{R}^{2}$,

For circular disc about its diameter ${I}_{3} = \dfrac{1}{4}M{R}^{2}$

And for circular ring about its diameter ${I}_{4} = \dfrac{1}{2}M{R}^{2}$

$⇒ 2\left(\dfrac{M{R}^{2}}{2} + \dfrac{M{R}^{2}}{4}\right) + \dfrac{M{R}^{2}}{2} = x \times \dfrac{2}{5}M{R}^{2}$

$⇒ 2\left(\dfrac{3}{4}\right) + \dfrac{1}{2} = x \times \dfrac{2}{5}$

$⇒ 2 = x \times \dfrac{2}{5}$

$⇒ x = 5$