A circular coil with 50 turns and a radius of 2.0 cm carries a 5.0...

The torque acting on a current-carrying coil in a magnetic field is given by $\tau = N I A B \sin \theta$

where $\theta$ is the angle between the area vector (normal to the plane of the coil) and the magnetic field.

The maximum torque is \tau_{\text\max} = N I A B

Given that the torque is half of its maximum possible value:

\tau = \dfrac{1}{2} \tau_{\text\max}

$N I A B \sin \theta = \dfrac{1}{2} N I A B$

$\sin \theta = \dfrac{1}{2}$

\theta = 30^{}^{{}^{\circ}}

The angle between the plane of the coil and the magnetic field is \alpha = 90^{}^{{}^{\circ}} - \theta

\alpha = 90^{}^{{}^{\circ}} - 30^{}^{{}^{\circ}} = 60^{}^{{}^{\circ}}