1. Use conservation of mechanical energy

For a body rolling without slipping from height $h$:

$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2$

Since rolling without slipping,

$v = \omega R \quad \Rightarrow \quad \omega = \frac{v}{R}$

So,

$Mgh = \frac{1}{2}Mv^2 + \frac{1}{2}I\frac{v^2}{R^2}$

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

Hence,

$v^2 = \frac{2Mgh}{M + I/R^2}$

2. For the solid cylinder

Moment of inertia about its axis:

$I_c = \frac{1}{2}MR^2$

Therefore,

$v_c^2 = \frac{2Mgh}{M + \frac{1}{2}M} = \frac{2Mgh}{\frac{3}{2}M} = \frac{4gh}{3}$

So,

$v_c = \sqrt{\frac{4gh}{3}}$

3. For the solid sphere

Moment of inertia about its axis:

$I_s = \frac{2}{5}MR^2$

Therefore,

$v_s^2 = \frac{2Mgh}{M + \frac{2}{5}M} = \frac{2Mgh}{\frac{7}{5}M} = \frac{10gh}{7}$

So,

$v_s = \sqrt{\frac{10gh}{7}}$

4. Find the ratio

$\frac{v_c}{v_s} = \sqrt{\frac{\frac{4gh}{3}}{\frac{10gh}{7}}}$

$\frac{v_c}{v_s} = \sqrt{\frac{4}{3}\cdot\frac{7}{10}} = \sqrt{\frac{28}{30}} = \sqrt{\frac{14}{15}}$

5. Match with options

$\boxed{\frac{v_c}{v_s} = \sqrt{\frac{14}{15}}}$

So the correct option is D.