1. Acceleration on the smooth incline

For a smooth incline at angle $45^\circ$, the acceleration is

$a_s = g\sin 45^\circ = \frac{g}{\sqrt{2}}$

Since the block starts from rest and slides distance $d$,

$d = \frac{1}{2} a_s t_s^2$

where $t_s$ is the time on the smooth incline.

2. Acceleration on the rough incline

On the rough incline, kinetic friction opposes motion.

Normal reaction:

$N = mg\cos 45^\circ$

Friction force:

$f_k = \mu_k N = \mu_k mg\cos 45^\circ$

Net acceleration down the incline:

$a_r = g\sin 45^\circ - \mu_k g\cos 45^\circ$

Since $\sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}}$,

$a_r = \frac{g}{\sqrt{2}}(1-\mu_k)$

3. Use the time relation

Given: time on rough incline is $n$ times the time on smooth incline.

$t_r = n t_s$

For the same distance $d$ from rest,

$d = \frac{1}{2} a_r t_r^2$

and also

$d = \frac{1}{2} a_s t_s^2$

Equating,

$a_s t_s^2 = a_r t_r^2$

Substitute $t_r = n t_s$:

$a_s t_s^2 = a_r n^2 t_s^2$

$a_s = a_r n^2$

So,

$a_r = \frac{a_s}{n^2}$

Now substitute expressions for $a_s$ and $a_r$:

$\frac{g}{\sqrt{2}}(1-\mu_k) = \frac{1}{n^2}\cdot \frac{g}{\sqrt{2}}$

Cancel $\frac{g}{\sqrt{2}}$:

$1-\mu_k = \frac{1}{n^2}$

Hence,

$\mu_k = 1 - \frac{1}{n^2}$

4. Match with options

Thus the correct option is

$\boxed{\mu_k = 1 - \frac{1}{n^2}}$

So, Option B is correct.

5. Comparison with stored answer

Stored correct answer: B

Our derived answer also gives B, so they agree.