1. Use energy conservation for rolling without slipping

For a body rolling down an incline 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,

$v^2 = \frac{2gh}{1+\frac{I}{mR^2}}$

2. Velocity of the circular ring

For a ring,

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

Thus,

$v_{\text{ring}}^2 = \frac{2gh}{1+1} = gh$

So,

$v_{\text{ring}} = \sqrt{gh}$

3. Velocity of the solid sphere

For a solid sphere,

$I_{\text{sphere}} = \frac{2}{5}mR^2$

Thus,

$v_{\text{sphere}}^2 = \frac{2gh}{1+\frac{2}{5}} = \frac{2gh}{\frac{7}{5}} = \frac{10gh}{7}$

So,

$v_{\text{sphere}} = \sqrt{\frac{10gh}{7}}$

4. Find the ratio

Taking the ratio of ring to sphere velocities:

$\frac{v_{\text{ring}}}{v_{\text{sphere}}} = \sqrt{\frac{gh}{10gh/7}} = \sqrt{\frac{7}{10}}$

Given,

$\sqrt{\frac{x}{5}} = \sqrt{\frac{7}{10}}$

Squaring both sides:

$\frac{x}{5} = \frac{7}{10}$

$x = \frac{7}{2} = 3.5$

5. Final answer

$\boxed{x = \frac{7}{2}}$

If integer format is enforced, the problem statement is inconsistent, because the correct value is $3.5$.

6. Comparison with stored answer

Stored correct answer: $4 \text{ OR } 3.5$

Our derived answer is $3.5$, which matches one of the accepted stored answers.