The cylinder rolls

Accleleration due to kinetic friction $a_f=\dfrac{f}{M} = \mu g\sin\theta = \dfrac{g\sqrt{3}}{2\sqrt{3}} = \dfrac{g}{2}$ and $g\sin\theta = \dfrac{g}{2}$ . So, the cylinder will roll until it stops and then start pure rolling.

$\begin{aligned}\alpha &= \dfrac{a_fR}{I}\\&=\dfrac{g}{R}\\\Rightarrow t_1 &= \dfrac{\omega_0R}{g}\end{aligned}$

Now, after this, it will start pure rolling with $v_0 = \omega_0 = 0$

Let $a_f$ be the acceleration due to static friction from this stage. $\begin{aligned}\Rightarrow a_{net} = R\alpha &= \dfrac{a_fR^2}{I}\\&= 2a_f\\ \text{Now, } g\sin\theta - a_f &= 2a_f\\\Rightarrow a&= \frac{2}{3}g\sin\theta \\&=\frac{1}{3}g\\\Rightarrow t_2 &= \sqrt{\dfrac{2s}{a_{net}}}\\&=\sqrt{\dfrac{2\times3R\cdot3}{g\sin\theta}}\\&=6\sqrt{\dfrac{R}{g}} \end{aligned}$

$\Large \therefore t = t_1 + t_2 =\boxed{ \dfrac{w_0R}{g}+6\sqrt{\dfrac{R}{g}}}$

BTW:This question is there in Cengage Textbook