Referring to above diagram,

balancing forces along the inclined plane,

$ma=mg\sin \left(\theta \right)-f ...\left(1\right)$

torque is given by,

$\tau =I\alpha =fR$ but $\alpha =\frac{a}{R} \& l=m{k}^2$

$\therefore m{k}^2\times \frac{a}{R}=fR\Rightarrow f=\frac{am{k}^2}{R^2}$

Putting in equation $\left(1\right)$,

$ma=mg\sin \left(\theta \right)-\frac{am{k}^2}{R^2}$

$\therefore a\left(1+\frac{k^2}{R^2}\right)=g\sin \left(\theta \right)$

$\therefore a=\frac{g\sin \left(\theta \right)}{\left(1+\frac{k^2}{R^2}\right)}$