At the topmost point of the loop minimum value of
the linear speed of the centre of the sphere should be :

                       $\mathit{\text{v}}=\sqrt{\mathit{\text{gR}}}$

or       translational kinetic energy

                   ${\mathit{\text{K}}}_{\text{T}}=\frac{1}{2}{\mathit{\text{mv}}}^2=\frac{1}{2}\mathit{\text{mgR}}\text{.}$

In case of pure rolling of a solid sphere the ratio
of rotational to translational kinetic energy is :

                                  $\frac{{\mathit{\text{K}}}_{\text{R}}}{{\mathit{\text{K}}}_{\text{T}}}=\frac{2}{5}$

∴ Total kinetic energy at topmost point should be :

              $\mathit{\text{K}}=\frac{5+2}{5}\text{.}{\mathit{\text{K}}}_{\text{T}}=\frac{7}{5}\left(\frac{1}{2}\mathit{\text{mgR}}\right)=\frac{7}{10}\mathit{\text{mgR}}$

Now from conservation of mechanical energy :

                     $\frac{7}{10}\mathit{\text{mgR}}=\mathit{\text{mg}}\left(\mathit{\text{h}}\text{ - 2}\mathit{\text{R}}\right)$

∴                       $h = 2.7 R$