b
As we Known, \(dU = F.dr\)

\(\begin{gathered}
  U = \int\limits_0^r {\alpha {r^2}} dr = \frac{{a{r^3}}}{3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right) \hfill \\
  As,\,\,\frac{{m{v^2}}}{r} = \alpha {r^2} \hfill \\
   \Rightarrow \,\frac{1}{2}m{v^2} = \frac{{\alpha \,{r^3}}}{2} \hfill \\
  or,\,\,\left( {KE} \right) = \frac{1}{2}\,\alpha {r^3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right) \hfill \\ 
\end{gathered} \)

Total energy = Potential energy + kinetic energy

Now, from eqn \((i)\) and \((ii)\) 

Total energy \(= K.E. + P.E.\)

\( = \frac{{\alpha {r^3}}}{3} + \frac{{\alpha {r^3}}}{2} = \frac{5}{6}\alpha {r^3}\)