A body rolls down an inclined plane  without slipping. The kinetic energy of rotation is 50% of its translational kinetic energy. The body is:

Translational kinetic energy $=\frac{1}{2}{\mathrm{mv}}^2$
Rotational kinetic energy $=\frac{1}{2}{\mathrm{I\omega }}^2$
According to question
$\begin{array}{l}\frac{1}{2}{\mathrm{I\omega }}^2=\frac{1}{2}\times \frac{1}{2}{\mathrm{mv}}^2\\ \text{ As v }=\mathrm{R\omega }\left(\text{ pure rolling }\right)\\ \frac{1}{2}{\mathrm{I\omega }}^2=\frac{1}{4}{\mathrm{mR}}^2{\mathrm{\omega }}^2\Rightarrow \mathrm{I}=\frac{1}{2}{\mathrm{mR}}^2\end{array}$
Thus, body should be solid cylinder.