1. Use energy conservation for rolling without slipping

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

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

Since rolling without slipping,

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

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 ring

For a ring,

$I_{\text{ring}} = mR^2$

Thus,

$v_r^2 = \frac{2mgh}{m + m} = \frac{2mgh}{2m} = gh$

So,

$v_r = \sqrt{gh}$

3. For the solid cylinder

For a solid cylinder,

$I_{\text{cyl}} = \frac{1}{2}mR^2$

Thus,

$v_c^2 = \frac{2mgh}{m + \frac{1}{2}m} = \frac{2mgh}{\frac{3}{2}m} = \frac{4gh}{3}$

So,

$v_c = \sqrt{\frac{4gh}{3}} = \frac{2}{\sqrt{3}}\sqrt{gh}$

4. Take the ratio

$\frac{v_r}{v_c} = \frac{\sqrt{gh}}{\frac{2}{\sqrt{3}}\sqrt{gh}} = \frac{\sqrt{3}}{2}$

Given,

$\frac{v_r}{v_c} = \frac{\sqrt{x}}{2}$

Therefore,

$\frac{\sqrt{x}}{2} = \frac{\sqrt{3}}{2}$

So,

$\sqrt{x} = \sqrt{3} \Rightarrow x = 3$

5. Final answer

$\boxed{3}$

The derived answer matches the stored correct answer.