Let the tension in the string attached to the wall be \(T\).

Resolve forces horizontally:

\(T \cos \alpha = 4 \sin 60^{\circ}\)

Resolve forces vertically:

\(T \sin \alpha + 4 \cos 60^{\circ} = 0.3g\)

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

\(T \sin \alpha + 4 \times 0.5 = 0.3 \times 9.8\)

\(T \sin \alpha = 2.94 - 2\)

\(T \sin \alpha = 0.94\)

Using the horizontal equation:

\(T \cos \alpha = 4 \times \frac{\sqrt{3}}{2}\)

\(T \cos \alpha = 2\sqrt{3}\)

Divide the two equations:

\(\tan \alpha = \frac{0.94}{2\sqrt{3}}\)

\(\alpha = \arctan \left( \frac{0.94}{2\sqrt{3}} \right) \approx 76.9^{\circ}\)

Substitute \(\alpha\) back to find \(T\):

\(T = \frac{2\sqrt{3}}{\cos 76.9^{\circ}} \approx 2.05 \text{ N}\)