Radius of the disc, $r=15 \,cm =0.15\, m$

The torsional oscillations of the disc has a time period, $T=1.5\, s$

The moment of inertia of the disc is:

${I}=\frac{1}{2} m r^{2}$

$={2} \times(10) \times(0.15)^{2}$

$=0.1125 \,kg\, m ^{2}$

$T=2 \pi \sqrt{\frac{I}{\alpha}} a$

Time period,

is the torsional constant.

$\alpha=\frac{4 \pi^{2} I}{T^{2}}$

$=\frac{4 \times(\pi)^{2} \times 0.1125}{(1.5)^{2}}$

$=1.972\, Nm / rad$

Hence, the torsional spring constant of the wire is $1.972\, Nm$ $rad$ $^{-1}$.