The correct option is (B) oscillations perpendicular to the plane of the hop.
The time period T for a physical pendulum is defined as,
T = 2π√I/Mgd
Here, I is the rotational inertia of the body about an axis through the pivot, d is the distance from the pivot to the center of mass, M is the mass the body and g is the acceleration due to gravity.
Here a round metal hoop is suspended on the edge by a hook. The hoop can oscillate side to side in the plane of the hoop, or it can oscillate back and forth in a direction perpendicular to the plane of the hoop.
Using equation T = 2π√I/Mgd, the angular frequency w of the oscillation would be,
w = 2πf
 =2π/T (Since, f = 1/T)
 =2π/(2π√I/Mgd)
= √Mgd/I
So the angular frequency w of the oscillation is inversely proportional to the rotational inertia of the body. Since rotational inertia of the hook when it oscillates in the plane of the hoop is greater than when it oscillates perpendicular to the plane of the hoop, thus the frequency of oscillation will be larger when the oscillations perpendicular to the plane of the hoop. Therefore option (B) is correct.