Correct Answer - B
The cylinder is having a fixed axis, so it can perform rotational motion only.
`fR=(MR^(2))/2alphaimpliesf=(MR)/2alpha`
For no slipping,
`Ralpha=` acceleration of the platform
`(2f)/M=omega^(2)Acosomegat`
For the maximum torque `f` would maximum and `f_(max)=Momega^(2)A/2`
So the maximum torque,
`tau_(max)=f_(max)xxR-=(M^(2)omega^(2)AR)/2`