Three objects, A: (a solid sphere), B: (a thin circular disk) and C = (a circular ring), each have the same mass M and radius R. They all spin with the same angular speed ω about their own symmetry axes. The amounts of work (W) required to bring them to rest, would satisfy the relation
1- \[{{w}_{c}}>{{w}_{b}}>{{w}_{a}}\]
2- \[{{w}_{a}}>{{w}_{b}}>{{w}_{c}}\]
3- \[{{w}_{b}}>{{w}_{a}}>{{w}_{c}}\]
4- \[{{w}_{a}}>{{w}_{c}}>{{w}_{b}}\]

Hint: All the three bodies are rotating. They all have the same mass and same radius. They are spinning at the same angular speed. The axis of rotation is the symmetry axis. We need to find the amount of work needed to be done to bring them at rest. We can use kinetic energy to solve this problem.

Complete step by step answer:
For a body rotating, kinetic energy is given by the formula, \[K=\dfrac{I{{\omega }^{2}}}{2}\], where I is the moment of inertia and \[\omega \]is the angular velocity. From work energy theorem work done is equal to change in kinetic energy. When the body stops, its final kinetic energy becomes zero. Here, the mass and radius for all the three bodies are the same M and r.
Moment of inertia for solid sphere is \[\dfrac{2M{{r}^{2}}}{5}\], for circular disk it is \[\dfrac{M{{r}^{2}}}{2}\]and for circular ring it is \[M{{r}^{2}}\].
So, kinetic energies of the three bodies comes out to be
For solid sphere(A)=\[\dfrac{2M{{r}^{2}}{{\omega }^{2}}}{5}\]
For circular disk(B)= \[\dfrac{M{{r}^{2}}{{\omega }^{2}}}{2}\]
For circular ring(C)=\[\dfrac{M{{r}^{2}}{{\omega }^{2}}}{2}\]

Thus, it is clear that \[{{w}_{c}}>{{w}_{b}}>{{w}_{a}}\]. So, the correct option is (1).

Note:
Here the three bodies were given and the moment of inertia is calculated about an axis passing through its centre of mass. So, the point to be kept in mind is the axis of rotation. Otherwise it is a simple and lucid problem. Work comes out in units of Joule, J.