First, resolve the tension in the string into horizontal and vertical components. The horizontal component is \(T \cos 15^{\circ} = 2.5 \cos 15^{\circ}\) and the vertical component is \(T \sin 15^{\circ} = 2.5 \sin 15^{\circ}\).

The normal reaction \(R\) is equal to the horizontal component of the tension: \(R = 2.5 \cos 15^{\circ}\).

The frictional force \(F\) is given by \(F = \mu R\). Therefore, \(F = \mu \times 2.5 \cos 15^{\circ}\).

In limiting equilibrium, the vertical forces balance: \(2.5 \sin 15^{\circ} = 0.03g + F\).

Substitute \(F = \mu \times 2.5 \cos 15^{\circ}\) into the equation: \(2.5 \sin 15^{\circ} = 0.03g + \mu \times 2.5 \cos 15^{\circ}\).

Rearrange to solve for \(\mu\):

\(\mu = \frac{2.5 \sin 15^{\circ} - 0.03g}{2.5 \cos 15^{\circ}}\).

Calculate \(\mu\) using \(g = 9.8\):

\(\mu = \frac{2.5 \times 0.2588 - 0.03 \times 9.8}{2.5 \times 0.9659} \approx 0.144\).