Resolve the forces horizontally:

\(20 \cos 60^\circ = T \cos 45^\circ\)

\(T = \frac{20 \cos 60^\circ}{\cos 45^\circ} = 10\sqrt{2}\) or 14.1 \text{ N}

Resolve the forces vertically:

\(20 \sin 60^\circ + T \sin 45^\circ = mg\)

\(m = \frac{20 \sin 60^\circ + T \sin 45^\circ}{g}\)

Substitute \(T = 10\sqrt{2}\):

\(m = \frac{20 \times \frac{\sqrt{3}}{2} + 10\sqrt{2} \times \frac{\sqrt{2}}{2}}{g} = 2.73 \text{ kg}\)