Let the mass of the block be denoted by \(m\).

The normal reaction force \(R\) is given by:

\(R = mg + 50 \sin 20^{\circ}\)

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

\(F = 0.3(mg + 50 \sin 20^{\circ})\)

Since the block moves at constant speed, the horizontal component of the applied force equals the frictional force:

\(50 \cos 20^{\circ} = 0.3(mg + 50 \sin 20^{\circ})\)

Solving for \(m\):

\(50 \cos 20^{\circ} = 0.3mg + 0.3 \times 50 \sin 20^{\circ}\)

\(50 \cos 20^{\circ} - 0.3 \times 50 \sin 20^{\circ} = 0.3mg\)

\(m = \frac{50 \cos 20^{\circ} - 0.3 \times 50 \sin 20^{\circ}}{0.3g}\)

Using \(g = 9.8 \text{ m/s}^2\), calculate \(m\):

\(m \approx 14.0 \text{ kg}\)