Time taken by the body to reach the bottom when it rolls down on an inclined plane without slipping is given by

$t = \sqrt {{{2l\left( {1 + {{{k^2}} \over {{R^2}}}} \right)} \over {g\sin \theta }}}$

Since g is constant and $l$, R and sin $\theta$ are same for both

$\therefore$ ${{{t_d}} \over {{t_s}}} = {{\sqrt {1 + {{k_d^2} \over {{R^2}}}} } \over {\sqrt {1 + {{k_s^2} \over {{R^2}}}} }} = \sqrt {{{1 + {{{R^2}} \over {2{R^2}}}} \over {1 + {{2{R^2}} \over {5{R^2}}}}}}$

$\left(\because {{k_d} = {R \over {\sqrt 2 }},{k_s} = \sqrt {{2 \over 5}} R} \right)$

$\Rightarrow \sqrt {{3 \over 2} \times {5 \over 7}} = \sqrt {{{15} \over {14}}} \Rightarrow {t_d} \gt {t_s}$