In limiting equilibrium, the frictional force \(F\) is given by \(F = \mu R\), where \(\mu = 0.25\) is the coefficient of friction and \(R\) is the normal reaction.

The horizontal component of the tension \(T\) is \(T \cos \alpha = 0.96T\), which equals the frictional force \(F\). Thus, \(F = 0.96T\).

The vertical component of the tension is \(T \sin \alpha\), and the weight of the ring is \(0.2g\). The normal reaction \(R\) is given by:

\(R = 0.2g - T\sin \alpha\)

Given \(\cos \alpha = 0.96\), we find \(\sin \alpha\) using \(\sin^2 \alpha + \cos^2 \alpha = 1\):

\(\sin \alpha = \sqrt{1 - 0.96^2} = 0.28\)

Thus, \(R = 0.2 \times 9.8 - T \times 0.28 = 1.96 - 0.28T\).

Using \(F = \mu R\), we have:

\(0.96T = 0.25(1.96 - 0.28T)\)

Solving for \(T\):

\(0.96T = 0.49 - 0.07T\)

\(0.96T + 0.07T = 0.49\)

\(1.03T = 0.49\)

\(T = \frac{0.49}{1.03} \approx 0.4757\)

Rounding to three significant figures, \(T = 0.485\).