1. Acceleration when the disc slips

If the disc slides without rolling constraint, the acceleration down the incline is just that of a particle:

$a_1 = g\sin\theta$

Given incline length is $l$ and initial velocity is zero,

$l = \frac{1}{2} a_1 t^2 = \frac{1}{2} g\sin\theta\, t^2$

So,

$t^2 = \frac{2l}{g\sin\theta}$

2. Acceleration when the disc rolls without slipping

For rolling motion,

$a_2 = \frac{g\sin\theta}{1 + \frac{I}{mR^2}}$

For a solid circular disc,

$I = \frac{1}{2}mR^2$

Hence,

$a_2 = \frac{g\sin\theta}{1 + \frac{1}{2}} = \frac{g\sin\theta}{3/2} = \frac{2}{3}g\sin\theta$

If time taken in rolling is $t_r$, then

$l = \frac{1}{2} a_2 t_r^2 = \frac{1}{2}\left(\frac{2}{3}g\sin\theta\right)t_r^2$

So,

$t_r^2 = \frac{2l}{a_2} = \frac{2l}{(2/3)g\sin\theta} = \frac{3l}{g\sin\theta}$

Using

$t^2 = \frac{2l}{g\sin\theta}$

we get

$t_r^2 = \frac{3}{2} t^2$

Therefore,

$t_r = \sqrt{\frac{3}{2}}\, t = \left(\frac{3}{2}\right)^{1/2} t$

3. Compare with given form

Given rolling time is

$\left(\frac{\alpha}{2}\right)^{1/2} t$

So,

$\left(\frac{\alpha}{2}\right)^{1/2} = \left(\frac{3}{2}\right)^{1/2}$

Hence,

$\alpha = 3$

4. Final answer

$\boxed{3}$