Answer
The inductance of the coil is
$$L=0.124 \text{ H}.$$
Work Step by Step
We will use the fact $\tan\varphi=\dfrac{\omega L-\frac{1}{\omega C}}{R}$ to relate $L$ and $R$ to $\varphi$. We know that the voltage across the coil leads the current in it by $\varphi=+52.3^{\circ}$.
There is no capacitance in the circuit so $X_C=0$ and
$\tan\varphi=\dfrac{X_L}{R}$
$X_L=R\tan\varphi=48.0\cdot\tan52.3^{\circ}=62.1\Omega$.
$X_L=\omega L=2\pi fL$
and
$L=\dfrac{62.1\Omega}{2\pi\cdot80.0\text{ Hz}}=0.124 \text{ H}.$
