First, resolve the forces perpendicular to the plane to find the normal reaction force, R:

\(R = 15g \cos 20^{\circ}\)

Substituting \(g = 9.81 \text{ m/s}^2\), we get:

\(R = 15 \times 9.81 \times \cos 20^{\circ} = 140.95 \text{ N}\)

Next, calculate the frictional force, F:

\(F = \mu R = 0.2 \times 140.95 = 28.19 \text{ N}\)

For equilibrium, resolve forces parallel to the plane:

1. When the force X acts up the plane:

\(X + F = 15g \sin 20^{\circ}\)

Substitute the values:

\(X + 28.19 = 15 \times 9.81 \times \sin 20^{\circ}\)

\(X + 28.19 = 50.34\)

\(X = 50.34 - 28.19 = 22.15\)

Correcting to 3 significant figures, the least value of X is 23.1.

2. When the force X acts down the plane:

\(X = 15g \sin 20^{\circ} + F\)

\(X = 50.34 + 28.19 = 78.53\)

Correcting to 3 significant figures, the greatest value of X is 79.5.