To solve this problem, we need to resolve the forces acting on the ring R both vertically and horizontally.

Vertically: The vertical components of the tension must balance the weight of the ring. The weight of the ring is given by:

\(0.2g\)

where \(g\) is the acceleration due to gravity (approximately 9.8 m/s²). The vertical components of the tension are:

\(T \sin 70^{\circ} + T \sin 45^{\circ} = 0.2g\)

Solving for \(T\):

\(T(\sin 70^{\circ} + \sin 45^{\circ}) = 0.2 \times 9.8\)

\(T = \frac{0.2 \times 9.8}{\sin 70^{\circ} + \sin 45^{\circ}} \approx 1.21 \text{ N}\)

Horizontally: The horizontal components of the tension must balance the horizontal force \(P\):

\(P + T \cos 70^{\circ} = T \cos 45^{\circ}\)

Substituting the value of \(T\):

\(P + 1.21 \cos 70^{\circ} = 1.21 \cos 45^{\circ}\)

Solving for \(P\):

\(P = 1.21 \cos 45^{\circ} - 1.21 \cos 70^{\circ} \approx 0.443 \text{ N}\)