A square is inscribed in a circle of radius $6$. A quarter circle is inscribed in the square, as shown in the diagram below. Given the area of the region inside the circle but outside the quarter circle is $n\pi$ for some positive integer $n$, what is $n$?
[asy]
        size(5 cm);
        draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);
        draw(circle((1,1),1.41));
        draw(arc((0,0),2,0,90));[/asy]


[i]2021 CCA Math Bonanza Lightning Round #1.2[/i]
To solve the problem, we need to find the area of the region inside the circle but outside the quarter circle. We will break this problem into several steps:

1. **Calculate the area of the circle:**
   The radius of the circle is given as \(6\). The area \(A\) of a circle is given by the formula:
   \[
   A = \pi r^2
   \]
   Substituting \(r = 6\):
   \[
   A = \pi \times 6^2 = 36\pi
   \]

2. **Determine the side length of the inscribed square:**
   Since the square is inscribed in the circle, the diagonal of the square is equal to the diameter of the circle. The diameter \(d\) of the circle is:
   \[
   d = 2 \times 6 = 12
   \]
   For a square with side length \(s\), the diagonal \(d\) is related to the side length by the Pythagorean theorem:
   \[
   d = s\sqrt{2}
   \]
   Solving for \(s\):
   \[
   s\sqrt{2} = 12 \implies s = \frac{12}{\sqrt{2}} = 6\sqrt{2}
   \]

3. **Calculate the area of the square:**
   The area \(A\) of the square is:
   \[
   A = s^2 = (6\sqrt{2})^2 = 36 \times 2 = 72
   \]

4. **Calculate the area of the quarter circle inscribed in the square:**
   The radius of the quarter circle is equal to the side length of the square, which is \(6\sqrt{2}\). The area \(A\) of a quarter circle is:
   \[
   A = \frac{1}{4} \pi r^2
   \]
   Substituting \(r = 6\sqrt{2}\):
   \[
   A = \frac{1}{4} \pi (6\sqrt{2})^2 = \frac{1}{4} \pi \times 72 = 18\pi
   \]

5. **Calculate the area of the region inside the circle but outside the quarter circle:**
   The area of the region inside the circle but outside the quarter circle is:
   \[
   \text{Area} = \text{Area of the circle} - \text{Area of the quarter circle}
   \]
   Substituting the areas calculated:
   \[
   \text{Area} = 36\pi - 18\pi = 18\pi
   \]

Thus, the value of \(n\) is \(18\).

The final answer is \(\boxed{18}\).