C

To solve for the speed of the particle at a distance \(x\) from the center of the ring, we first recognize the gravitational potential energy at that position. The gravitational force on the particle due to the ring provides the acceleration needed to find the speed as it moves from rest under this force.
1. **Gravitational Potential Energy (U)**: Gravitational potential energy at distance \(x\) from the center is given by:
\[ U = -\frac{GMm}{\sqrt{r^2 + x^2}} \]
where \(M\) is the mass of the ring and \(m\) is the mass of the particle.
2. **Kinetic Energy (K)**: When the particle reaches speed \(v\):
\[ K = \frac{1}{2} mv^2 \]
3. **Conservation of Energy**:
Initial potential energy at height, which is zero (since the particle starts from rest), and as it falls, all this potential energy converts to kinetic energy:
\[ 0 = U + K \implies 0 = -\frac{GMm}{\sqrt{r^2 + x^2}} + \frac{1}{2} mv^2 \]
4. **Solving for Speed**: Rearranging gives us:
\[ \frac{1}{2} mv^2 = \frac{GMm}{\sqrt{r^2 + x^2}} \implies v^2 = \frac{2GM}{\sqrt{r^2 + x^2}} \]
Therefore, \(v = \sqrt{\frac{2GM}{\sqrt{r^2 + x^2}}} \).
5. For specific values substituting in gives the correct choice from the options. This computation confirms that option C is the correct choice.