This question will be faulty until you give some relation between radius of curvatures of pt. A,B and C.

Now,if radius of curvatures at all 3 pts are assumed to be equal,then:

For A,

      N+(mv^2/r)=mg;           because centrifugal for will act along the radial vector joining A and centre of curvature and in opposite           N=mg-(mv^2/r)             direction,so will get added to normal reaction force(N) which is along the same direction.

For B,

      N=mg+(mv^2/r);           because N will act in up direction while centrifugal force and weight will act in forward direction.

For C,

      Same as A.

Therefore,N(A)=N(C)<N(B).

NOTE:This question has been solved assuming the plane of motion to be vertical.

          If the plane of motion is horizontal,N(A)=N(B)=N(C)=mg.

IS THE ANSWER N2>N1>N3?

in this question if you consider your reference frame on the car and then apply the concept of centrifugal force.

now,at point A,

mg=n+centrifugal

at B,

mg+centrifugal=n

so the normal at b is definately more than at a.for c we have to note that the centrifugal depends on radius and hence more the radius less the centrifugal.