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

Since g is constant and I, R and sin \(\theta \) are same for both

\({{{t_d}}}{{{t_s}}} = \frac{{\sqrt {1 + \frac{{k_d^2}}{{{R^2}}}} }}{{\sqrt {1 + \frac{{k_s^2}}{{{R^2}}}} }} = \sqrt {\frac{{1 + \frac{{{R^2}}}{{2{R^2}}}}}{{1 + \frac{{2{R^2}}}{{5{R^2}}}}}}\)
\({{k_d} = \frac{R}{{\sqrt 2 }},{k_s} = \sqrt {\frac{2}{5}} R}\) 
\(= \sqrt {\frac{3}{2} \times \frac{5}{7}}  = \sqrt {\frac{{15}}{{14}}}  \Rightarrow {t_d} > {t_s}\)

Hence, the sphere gets to the bottom first.