If $k < (\sqrt{2}+1)^3 < k+2$, where $k$ is a natural number, then what is the value of $k$?

Let $x = (\sqrt{2}+1)^3$.

Let $x = I + f$, where $I$ is the integer part and $0 < f < 1$ is the fractional part.

Let $f' = (\sqrt{2}-1)^3$. Since $0 < \sqrt{2}-1 < 1$, we have $0 < f' < 1$.

Now, evaluate $x - f'$:

$x - f' = (\sqrt{2}+1)^3 - (\sqrt{2}-1)^3$

$x - f' = (2\sqrt{2} + 6 + 3\sqrt{2} + 1) - (2\sqrt{2} - 6 + 3\sqrt{2} - 1)$

$x - f' = (7 + 5\sqrt{2}) - (5\sqrt{2} - 7) = 14$

Since $x = I + f$, we get $I + f - f' = 14$, which implies $f - f' = 14 - I$.

Since $14 - I$ is an integer and $-1 < f - f' < 1$, the only possible integer value for $f - f'$ is $0$.

Thus, $f = f'$ and $I = 14$.

This gives $x = 14 + f$, meaning $14 < x < 15$.

We are given the inequality $k < x < k+2$.

Checking the given options:

For $k = 11$, the interval is $(11, 13)$, which does not contain $x$.

For $k = 13$, the interval is $(13, 15)$, which contains $x$ since $14 < x < 15$.

For $k = 15$, the interval is $(15, 17)$, which does not contain $x$.

For $k = 17$, the interval is $(17, 19)$, which does not contain $x$.

Thus, the correct value of $k$ from the options is $13$.

Answer: $13$