To solve this problem, we resolve the forces acting on the ring R in both horizontal and vertical directions.

1. Resolve horizontally: The horizontal component of the tension T in the string is balanced by the horizontal component of the 2.5 N force.

\(T \cos 45^{\circ} + T \cos 45^{\circ} = 2.5 \cos 45^{\circ}\)

\(2T \cos 45^{\circ} = 2.5 \cos 45^{\circ}\)

\(T = 1.25 \text{ N}\)

2. Resolve vertically: The vertical component of the tension balances the weight of the ring.

\(2.5 \sin 45^{\circ} = mg\)

\(m = \frac{2.5 \sin 45^{\circ}}{g}\)

Using \(g = 9.8 \text{ m/s}^2\),

\(m = \frac{2.5 \times \frac{\sqrt{2}}{2}}{9.8} \approx 0.177 \text{ kg}\)

Thus, the tension in the string is 1.25 N and the mass of the ring is 0.177 kg.