It can be shown that: $$T = 2\pi\sqrt{\frac{l}{c}}$$ where c = Couple per unit twist
But, $$c = \frac{\pi\eta\alpha^4}{2l}$$ And, $$\eta = \frac{\text{shearing stress}}{\text{Strain(shear angle)}} = \frac{8\pi l}{\alpha^4} * \frac{l}{T^2}$$ here, I is the moment of inertial of the disc.
Finally we get, $$l = \left(\frac{\eta \alpha^4}{4\pi MR^2}\right) T^2$$ The equation above can be plotted l along y-axis and T 2 along x-axis and from the slope the Rigidity modulus can be found out.