To solve this problem, we need to compare the time taken for the block to slide down the inclined plane with and without friction.

Without Friction:

When there is no friction, the only force acting along the inclined plane is the component of the gravitational force. The acceleration $a_{\text{nf}}$ (where "nf" stands for no friction) can be given by:

$$ a_{\text{nf}} = g \sin(\theta) $$

Where:

$g$ is the acceleration due to gravity.

$\theta = 45^\circ$, so $\sin(45^\circ) = \frac{\sqrt{2}}{2}$.

Thus:

$$ a_{\text{nf}} = g \frac{\sqrt{2}}{2} $$

With Friction:

When friction is present, the net force along the plane is reduced by the frictional force. The acceleration $a_{\text{f}}$ (where "f" stands for friction) is:

$$ a_{\text{f}} = g \sin(\theta) - \mu g \cos(\theta) $$

Where:

$\mu$ is the coefficient of friction.

$\cos(45^\circ) = \frac{\sqrt{2}}{2}$.

Thus:

$$ a_{\text{f}} = g \frac{\sqrt{2}}{2} - \mu g \frac{\sqrt{2}}{2} $$

$$ a_{\text{f}} = g \frac{\sqrt{2}}{2} (1 - \mu) $$

Relation between Times:

From the problem, the time with friction $t_{\text{f}}$ is three times the time without friction $t_{\text{nf}}$. Thus:

$$ t_{\text{f}} = 3 t_{\text{nf}} $$

Since time $t$ under constant acceleration is inversely proportional to the square root of the acceleration:

$$ t \propto \frac{1}{\sqrt{a}} $$

Therefore:

$$ \frac{1}{\sqrt{a_{\text{f}}}} = 3 \cdot \frac{1}{\sqrt{a_{\text{nf}}}} $$

Squaring both sides gives:

$$ \frac{1}{a_{\text{f}}} = 9 \cdot \frac{1}{a_{\text{nf}}} $$

Hence:

$$ a_{\text{f}} = \frac{a_{\text{nf}}}{9} $$

Substitute for $a_{\text{f}}$ and $a_{\text{nf}}$:

$$ g \frac{\sqrt{2}}{2} (1 - \mu) = \frac{g \frac{\sqrt{2}}{2}}{9} $$

Canceling terms common on both sides gives:

$$ 1 - \mu = \frac{1}{9} $$

Solving for $\mu$:

$$ \mu = 1 - \frac{1}{9} $$

$$ \mu = \frac{8}{9} $$

Therefore, the coefficient of friction is approximately 0.88. So, the correct option is:

Option C: 0.88

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