Hold your ground.

The tendency of jumping of hoop will be maximum when the body A is at top most position.

Let at this position the linear velocity of hoop is $V$ . So, velocity of the body is $2V$ .

Now for hoop to jump, Centrifugal force given by the body on hoop must be greater than weight of the combined system.

So, $F_{cp}=\frac{m{V}^2}{R}=2mg$ (Here in $F_{cp}$ only $V$ is taken into account and not $2V$ because the velocity of body A relative to the centre of mass of hoop is $V$ )

or, $V=\sqrt{2gR}$

Since there is no energy loss, so we can use conservation of energy.

So, $\frac{1}{2}I_{cm}$ ${w_{i}}^{2}$ + $\frac{1}{2}(2m){{v_{cm}}_{i}}^{2}$ = $(2m)gR$ + $\frac{1}{2}I_{cm}$ ${w_{f}}^{2}$ + $\frac{1}{2}(2m){{v_{cm}}_{f}}^{2}$

Therefore $\frac{1}{2}\frac{3m{R}^{2}{V_{o}}^{2}}{2{R}^{2}}$ + $\frac{m{{V_{o}}^{2}}}{4}$ = $2mgR$ + $\frac{1}{2}\frac{3m{R}^{2}{V}^{2}}{2{R}^{2}}$ + $m{(\frac{3V}{2})}^{2}$ [Here ${v_{cm}}_{i}$ = $\frac{V_{o}}{2}$ & ${v_{cm}}_{f}$ = $\frac{3V}{2}$ ]

or, $m{V_{o}}^{2}$ = $2mgR$ + $3m{V}^{2}$

or, ${V_{o}}^{2}$ = $2gR$ + $3{V}^{2}$

Now putting the value of $V=\sqrt{2gR}$ in above equation

${V_{o}}^{2}$ = $2gR$ + $3(2gR)$

or, ${V_{o}}^{2}$ = $8gR$

or, $V_{o}$ = $\sqrt{8gR}$ .

So to prevent hoop from jumping $V_{o}$ must be less than or equal to $\sqrt{8gR}$ .

The tendency of jumping of hoop will be maximum when the body A is at top most position.

Let at this position the linear velocity of hoop is . So, velocity of the body is .

Now for hoop to jump, Centrifugal force given by the body on hoop must be greater than weight of the combined system.

So, (Here in only is taken into account and not because the velocity of body A relative to the centre of mass of hoop is )

or,

Since there is no energy loss, so we can use