Analysis:
To determine the coefficient of friction between the plane and the ring, we can use the concept of rolling without slipping on an inclined plane. When an object is rolling without slipping on an inclined plane, the coefficient of friction between the object and the plane must be greater than a certain value to prevent slipping.

Key Points:
- The angle of the inclined plane is given as 45°.
- The ring is rolling without slipping on the inclined plane.
- We need to find the minimum coefficient of friction required for the ring to roll without slipping.

Solution:
Given that the ring is rolling without slipping on the inclined plane, we know that the frictional force must be sufficient to prevent slipping. The force of friction can be calculated using the formula:
\( f = \mu \times N \)
Where:
- \( f \) is the frictional force
- \( \mu \) is the coefficient of friction
- \( N \) is the normal force acting on the ring
The normal force can be resolved into two components:
- \( N \cos \theta \) perpendicular to the plane
- \( N \sin \theta \) parallel to the plane
For the ring to roll without slipping, the frictional force \( f \) must be equal to or greater than the force component parallel to the plane, which is given by:
\( f \geq N \sin \theta \)
Substitute the expression for \( f \) in terms of \( \mu \) and \( N \) into the inequality:
\( \mu \times N \geq N \sin \theta \)
\( \mu \geq \sin \theta \)
Solving for \( \theta = 45° \):
\( \mu \geq \sin 45° = \frac{1}{\sqrt{2}} \approx 0.707 \)
Therefore, the coefficient of friction between the plane and the ring must be greater than \( \frac{1}{\sqrt{2}} \) for the ring to roll without slipping on the inclined plane with an angle of 45°.

Conclusion:
The coefficient of friction between the plane and the ring must be greater than \( \frac{1}{\sqrt{2}} \) for the ring to roll without slipping on an inclined plane with an angle of 45°. This corresponds to option A) \( \frac{1}{2} \).