Explanation

(b) The acceleration of the body which is rolling down an inclined plane of angle $\alpha$ is
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$a = \frac{g\ sin\ \alpha}{1 + \frac{K_{2}}{R^{2}}}$
where $K =$radius of gyration,
$R =$radius of body.
Now, here the body is a uniform solid disc.
So, $\frac{K^{2}}{R^{2}} = \frac{1}{2}$
$\therefore$ $a = \frac{g\ sin\ \alpha}{1 + \frac{1}{2}}$
or $a = \frac{g\ sin\alpha}{\frac{3}{2}}$
or $a = \frac{2g\sin\alpha}{3}$