What we want to find is work done. We know that the work done to stop a moving body is the total kinetic energy of the body.

∴ Work Done = KE = KE_rot + KE_tran

Where KE_rot is rotational kinetic energy and

KE_tran is translational kinetic energy

KE_rot = 1/2 Iω²

KE_tran = 1/2 Mv²

Where M is mass of hoop

I is moment of inertia of hoop

v is velocity of centre of mass

ω is angular velocity

Now the given values are,

Mass of hoop, M = 100 Kg

Radius of hoop, R = 2 m

Speed of centre of mass of hoop, v = 20 Cm s^-1

= .2 m s^-1

The axis of rotation of hoop passes through point B, we know that velocity of point O (centre of loop) is v. If ω is angular velocity of hoop then,

v = R ω (R is Distance between axis and centre(O))

Where ω is angular velocity.

i.e. ω = v/R = 0.2/2 = .1 s^-1

Moment of inertia of hoop, I = MR²

Where, M is mass of hoop

R is Radius of hoop

∴ I = 100(2)²

I = 400 Kg m²

Total Kinetic Energy of hoop,

KE = KE_rot + KE_tran

KE_rot = 1/2 400 × (.1)²

= 2 J

KE_tran = 1/2 100 × (.2)²

= 2 J

∴ Total Kinetic energy,

KE = 2 + 2 = 4 J

Work done to stop the moving hoop is equal to its KE which is 4 J