Answer: C. $88$

Explanation:1. Understand the standard equation of a circle: The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.

2. Convert the given equation to the standard form by completing the square: The given equation is $x^2 - 4x + y^2 - 12y - 81 = 0$. Group x-terms and y-terms and move the constant to the right side: $(x^2 - 4x) + (y^2 - 12y) = 81$ To complete the square for $x^2 - 4x$, take half of the coefficient of x ($\dfrac{-4}{2} = -2$) and square it ($(-2)^2 = 4$). Add $4$ to both sides. $(x^2 - 4x + 4) + (y^2 - 12y) = 81 + 4$ To complete the square for $y^2 - 12y$, take half of the coefficient of y ($\dfrac{-12}{2} = -6$) and square it ($(-6)^2 = 36$). Add $36$ to both sides. $(x^2 - 4x + 4) + (y^2 - 12y + 36) = 81 + 4 + 36$ Rewrite in factored form: $(x - 2)^2 + (y - 6)^2 = 121$

3. Identify the radius of the circle: Comparing with $(x-h)^2 + (y-k)^2 = r^2$: Center $(h, k) = (2, 6)$. $r^2 = 121$. Radius $r = \sqrt{121} = 11$.

4. Relate the inscribed circle to the square: If a circle is inscribed in a square, the diameter of the circle is equal to the side length of the square. Diameter $d = 2r = 2 \times 11 = 22$. So, the side length of the square ($s\_{sq}$) is $22$.

5. Calculate the perimeter of the square: Perimeter of a square $P = 4 \times s\_{sq}$. $P = 4 \times 22 = 88$.

6. Conclusion: The perimeter of the square is $88$.