What is the power given by centripetal force?(in circular motion) 

A particle of mass $m$ is moving in a circular path of constant radius
  such that the centripetal acceleration is varying with time as $a_c =
 k^2rt^2$ where $k$ is constant. The power given to the particle by the
  force acting on it is?

MY ATTEMPT:
Since force and velocity are radial power must be zero, but the answer is not zero . If I find instantaneous power then I am getting $P_i = 2 \pi m k^2 r^2 t$.
Where $P_i$ = instantaneous power,
The answer given is option (b), as in the following figure, but why?
 A: You mean "the force is radial and the velocity is transversal". The power is the change of kinetic energy per time, dE/dt where E=m*v²/2. So it must be
$$P_i = m\cdot k^2\cdot r^2\cdot t$$
The factor $\pi$ or $2\pi$ is not nescessary since that would be a half circle while what you need is a radian and that is $1$ in natural units.
A: The particle is moving on a circle, but with variable velocity. So it s a non-uniform circular motion. You see this immediately because the centripetal acceleration is not constant. This means that there is some power given/taken to the particle. 
The text gives you the component of the acceleration along the radius of curvature (centripetal acceleration), but it does not say that the tangential acceleration is 0. In fact, it cannot be 0: there must be a tangential acceleration which increases the speed of the particle.
Anyway:
from $r.r = const$ (circular motion)
derive once and get $2 v.r= 0$, so $v$ is orthogonal to $r$ (as we concluded before).
Derive a second time $ v.r= 0$ and get $a.r+v.v=0$.
The first term is minus the centripetal acceleration (which is $-a.n_r$) times the length of $r$, so is $-k^2r^2t^2$ ($r$ is pointing outward, the centripetal acceleration must point toward $-r$ ie. inward and $n_r$ is a unit radial vector). 
So you have $-k^2r^2t^2+v.v=0$. Multiply by $1/2m$ and derive again and note that the second term is the derivative of the kinetic energy, ie. power received
$$d/dt(-1/2mk^2r^2t^2+1/2mv.v)=0$$
$$-mk^2r^2t+P=0$$
and you get answer b correctly
