A particle is moving in a circle of radius r under the action of a force $\mathrm{F}={\mathrm{\alpha r}}^2$ which is directed towards centre of the circle. Total mechanical energy (kinetic energy + potential energy) of the particle is (take potential energy = 0 for r = 0) :

The particle is being centripetal accelerated by the force F.
The  kinetic energy is $\frac{1}{2}{\mathrm{mv}}^2$

$$\begin{gathered} \mathrm{F}=\frac{{\mathrm{mv}}^2}{\mathrm{r}}={\mathrm{\alpha r}}^2 \\ \Rightarrow \frac{1}{2}{\mathrm{mv}}^2=\frac{{\mathrm{\alpha r}}^3}{2} \end{gathered}$$

Potential energy is the work that force F does when it moves the mass from r=0 to r=r.

$$\begin{gathered} \mathrm{U}={\int }_0^{\mathrm{r}}\overline{\mathrm{F}}\mathrm{d}\overline{\mathrm{r}}={\int }_0^{\mathrm{r}}{\mathrm{\alpha r}}^2\mathrm{dr} \\ \Rightarrow \mathrm{U}=\frac{{\mathrm{\alpha r}}^3}{3} \end{gathered}$$

Thus the total energy is = kinetic energy + potential energy

$\mathrm{Total} \mathrm{energy}=\frac{{\mathrm{\alpha r}}^3}{2}+\frac{{\mathrm{\alpha r}}^3}{3}=\frac{5{\mathrm{\alpha r}}^3}{6}.$

Hence the correct answer is $\frac{5{\mathrm{\alpha r}}^3}{6}.$