Question

A HOOP is suspended on a peg in a wall its moment of inertia about the peg is(mass=M and radius =R)

Answer 1

By the word suspended,I take that the ring (hoop) is put on a peg on a wall.
Using the perpendicular axis theorem,
    MI(peg) = MI(center of hoop) + (Mass of hoop)x(radius)²
                    MR² + MR² = 2MR²

Answer 2

Moment of Inertia of Hoop about the peg is 2M$R^{2}$.

Explanation:

A hoop is suspended on a peg in a wall .

A hoop is considered equivalent to a standard ring.

Mass Moment of Inertia of a ring about its center of mass is equal to

$I_{com} =MR^{2}$  

where M is mass of ring and R is radius of ring.

The hoop is suspended on a peg in a wall about a point on the circumference of ring.

Parallel Axis Theorem

Moment of  Inertia about a point "a" distance away from center of mass is equal to the sum of moment of inertia about COM and product of mass and square of the distance.

$I = I_{com} + Ma^{2}$

a = R for this situation;

$I = I_{com} + MR^{2} \\I = MR^{2} + MR^{2} \\I = 2MR^{2}$