A particle of mass m moves along a circle of radius R with a centripetal acceleration varying with time as a_e  = Kt² where K is a constant. Then

Since the particle is moving along a circle.
Normal acceleration, $\frac{{\mathrm{v}}^2}{\mathrm{R}}={\mathrm{kt}}^2\text{ or }\mathrm{v}=\mathrm{t}\sqrt{\mathrm{kR}}$
Tangential acceleration, ${\mathrm{a}}_{\mathrm{t}}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{t}\sqrt{\mathrm{kR}}\right)$
$\therefore$Tangential component of resultant force ${\mathrm{F}}_{\mathrm{t}}={\mathrm{ma}}_{\mathrm{t}}=\mathrm{m}\sqrt{\mathrm{kR}}$
Power developed by normal force  F_n is always zero because its direction is always perpendicular to the direction of motion of the particle So, power developed by tangential force alone
$\mathrm{P}=\overrightarrow{{\mathrm{F}}_{\mathrm{t}}}\cdot \overrightarrow{\mathrm{v}}=\mathrm{m}\sqrt{\mathrm{kR}}\cdot \mathrm{t}\sqrt{\mathrm{kR}}=\mathrm{mkRt}$
Power developed by resultant force is used to increase kinetic energy of the body.
Initial K.E at t = 0 , E₀  = 0
K.E at time $\mathrm{t},\mathrm{E}=\frac{1}{2}{\mathrm{mv}}^2=\frac{1}{2}{\mathrm{mkRt}}^2$
Average power developed = Average rate of increase of K.E
Average power $=\frac{\mathrm{E}-{\mathrm{E}}_0}{\mathrm{t}}=\frac{1}{2}\mathrm{mkRt}$