Current flowing in the coil, $I=12\, A$

Number of turns on the coil, $n=20$

Angle made by the plane of the coil with magnetic field, $\theta=30^{\circ}$

Strength of magnetic field, $B=0.80\, T$

Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,

$\tau=n B L A \sin \theta$

$\Rightarrow l \times l=0.1 \times 0.1=0.01\, m^{2}$

$\therefore \tau=20 \times 0.8 \times 12 \times 0.01 \times \sin 30^{\circ}$

$=0.96 \,Nm$

Hence, the magnitude of the torque experienced by the coil is $0.96 \,N m$