First, calculate the normal reaction force, \(R\), on the block. Since the block is on an inclined plane, \(R = 5g \cos 6^{\circ}\).

Next, calculate the frictional force, \(F\), using the formula \(F = \mu R\), where \(\mu = 0.3\). Thus, \(F = 0.3 \times 5g \cos 6^{\circ}\).

Since the block is moving at constant speed, the tension \(T\) in the rope must balance the component of gravitational force down the plane and the frictional force. Therefore, \(T = 5g \sin 6^{\circ} + F\).

Substitute the values to find \(T\):

\(T = 5g \sin 6^{\circ} + 0.3 \times 5g \cos 6^{\circ}\).

Calculating this gives \(T \approx 20.1 \text{ N}\).