The time taken to slide down the rough surface is twice that of the smooth surface: $t_{\text{rough}} = 2t_{\text{smooth}}$.

The acceleration on the smooth surface is given by:

$ a_{\text{smooth}} = g \sin \theta $

The time to slide down an inclined plane is inversely proportional to the square root of the acceleration:

$ t \propto \frac{1}{\sqrt{a}} \Rightarrow t_{\text{smooth}} \propto \frac{1}{\sqrt{g \sin \theta}} $

The acceleration on the rough surface is:

$ a_{\text{rough}} = g \sin \theta - \mu_k g \cos \theta $

Relating the times on both surfaces, we have:

$ \frac{t_{\text{rough}}}{t_{\text{smooth}}} = \frac{\sqrt{\sin \theta}}{\sqrt{\sin \theta - \mu_k \cos \theta}} = 2 $

Squaring both sides results in:

$ \frac{\sin \theta}{\sin \theta - \mu_k \cos \theta} = 4 $

Substituting $\theta = 45^{\circ}$ where $\sin \theta = \cos \theta = \frac{1}{\sqrt{2}}$, we find:

$ \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} - \mu_k \times \frac{1}{\sqrt{2}}} = 4 $

Simplifying further:

$ 1 - \mu_k = \frac{1}{4} $

Solving for $\mu_k$:

$ \mu_k = \frac{3}{4} = 0.75 $

Thus, the coefficient of kinetic friction is $0.75$.