The kinetic energy of the rolling object is converted into potential energy at height

$h = \left( { {{3{v^2}} \over {4g}}} \right)$

So by the law of conservation of mechanical energy, we have

${1 \over 2}M{v^2} + {1 \over 2}I{\omega ^2} = Mgh$  $\left( \because{\omega = {v \over R}} \right)$

${1 \over 2}M{v^2} + {1 \over 2}I{\left( {{v \over R}} \right)^2} = Mg\left( {{{3{v^2}} \over {4g}}} \right)$

${1 \over 2}I{{{v^2}} \over {{R^2}}} = {3 \over 4}M{v^2} - {1 \over 2}M{v^2}$

${1 \over 2}I{{{v^2}} \over {{R^2}}} = {1 \over 4}M{v^2}$

$\Rightarrow I = {1 \over 2}M{R^2}$

Hence, the object is disc.