To solve for the tension in the string and the value of X, we resolve the forces acting on the ring R.

1. Resolve vertically: \(X \sin 60^{\circ} + T \sin 30^{\circ} - 0.2g = 0\)

2. Resolve horizontally: \(X \cos 60^{\circ} - T - T \cos 30^{\circ} = 0\)

Solving these equations simultaneously:

From the vertical resolution, substitute the known values: \(X \cdot \frac{\sqrt{3}}{2} + T \cdot \frac{1}{2} = 0.2 \cdot 9.8\)

From the horizontal resolution: \(X \cdot \frac{1}{2} = T + T \cdot \frac{\sqrt{3}}{2}\)

Solving these equations gives: \(X = 2\) \(T = 0.536 \text{ N}\)