Lời giải

* \( 6x - x^2 = m \Leftrightarrow x^2 - 6x + m = 0 \)

\( \Leftrightarrow  \begin{cases} 0 < m < 9 \\ x = 3 \pm \sqrt{9 - m} \end{cases} \)

* \( \int_{0}^{6} \left( 6x - x^2 \right) \, dx =  3x^2 - \frac{x^3}{3} \bigg|_0^6 = 36 \)

Tìm \( m \) sao cho:  
\( \int_{3 - \sqrt{9 - m}}^{3 + \sqrt{9 - m}} \left( 6x - x^2 - m \right) \, dx = 18 \)

\( \Rightarrow \int_{3 - \sqrt{9 - m}}^{3 + \sqrt{9 - m}} \left[ (9 - m) - (x - 3)^2 \right] \, dx \)

\( =  (9 - m)x - \frac{(x - 3)^3}{3} \bigg|_{3 - \sqrt{9 - m}}^{3 + \sqrt{9 - m}} = \frac{4}{3} \left( \sqrt{9 - m} \right)^3 = 18 \)

\( \Rightarrow \left( \sqrt{9 - m} \right)^3 = \frac{27}{2} \quad \Rightarrow \sqrt{9 - m} = \frac{3}{\sqrt[3]{2}} \Leftrightarrow m = \frac{9 \cdot \sqrt[3]{4} - 9}{\sqrt[3]{4}}\)

\( \Leftrightarrow m = \frac{9 \left( 2 -  \sqrt[3]{2} \right)}{2} \)
 

\(\Rightarrow\) Vậy chọn đáp án \(\boxed{\text{D}} \)