To solve this problem, we need to resolve the forces acting on the block both perpendicular and parallel to the inclined plane.

Step 1: Resolve forces perpendicular to the plane.

The normal reaction force \(R\) and the component of the tension \(T \sin 20^{\circ}\) act perpendicular to the plane. The weight component \(2.5g \cos 30^{\circ}\) acts perpendicular to the plane as well.

Thus, the equation is:

\(R + T \sin 20^{\circ} = 2.5g \cos 30^{\circ}\)

Step 2: Friction force.

The friction force \(F\) is given by:

\(F = 0.25 \times R\)

Step 3: Resolve forces parallel to the plane.

The tension component \(T \cos 20^{\circ}\) acts up the plane, while the friction force \(F\) and the weight component \(2.5g \sin 30^{\circ}\) act down the plane.

Thus, the equation is:

\(T \cos 20^{\circ} = F + 2.5g \sin 30^{\circ}\)

Step 4: Solve the equations.

Substitute \(F = 0.25 \times R\) into the parallel force equation:

\(T \cos 20^{\circ} = 0.25R + 2.5g \sin 30^{\circ}\)

Using the perpendicular force equation, solve for \(R\):

\(R = 2.5g \cos 30^{\circ} - T \sin 20^{\circ}\)

Substitute \(R\) into the parallel force equation and solve for \(T\):

\(T \cos 20^{\circ} = 0.25(2.5g \cos 30^{\circ} - T \sin 20^{\circ}) + 2.5g \sin 30^{\circ}\)

Solving this equation gives \(T = 17.5\).