Đáp án đúng là: B

Ta có: \(3\sin 3{\rm{x}} - \sqrt 3 cos9{\rm{x}} = 1 + 4{\sin ^3}3{\rm{x}}\)

\( \Leftrightarrow \left( {3\sin 3{\rm{x}} - 4{{\sin }^3}3{\rm{x}}} \right) - \sqrt 3 cos9{\rm{x}} = 1\)

\( \Leftrightarrow \sin 9x - \sqrt 3 cos9{\rm{x}} = 1\)

\( \Leftrightarrow \frac{1}{2}\sin 9x - \frac{{\sqrt 3 }}{2}cos9{\rm{x}} = \frac{1}{2}\)

\( \Leftrightarrow cos\frac{\pi }{3}\sin 9x - \sin \frac{\pi }{3}cos9{\rm{x}} = \frac{1}{2}\)

\( \Leftrightarrow \sin \left( {9x - \frac{\pi }{3}} \right) = \frac{1}{2}\)

\( \Leftrightarrow \sin \left( {9x - \frac{\pi }{3}} \right) = \sin \frac{\pi }{6}\)

\( \Leftrightarrow \left[ \begin{array}{l}9{\rm{x}} - \frac{\pi }{3} = \frac{\pi }{6} + k2\pi \\9{\rm{x}} - \frac{\pi }{3} = \frac{\pi }{6} + k2\pi \end{array} \right.\)

\( \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{18}} + \frac{{k2\pi }}{9}\\x = \frac{{7\pi }}{{54}} + \frac{{k2\pi }}{9}\end{array} \right.\)

Vì x > 0 nên \(\left[ \begin{array}{l}\frac{\pi }{{18}} + \frac{{k2\pi }}{9} > 0\\\frac{{7\pi }}{{54}} + \frac{{k2\pi }}{9} > 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\pi + 4k\pi > 0\\7\pi + 12k\pi > 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}k > \frac{{ - 1}}{4}\\k > \frac{{ - 7}}{{12}}\end{array} \right.\)

\( \Rightarrow \left[ \begin{array}{l}{k_{\min }} = 0\\{k_{\min }} = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{{18}}\\x = \frac{{7\pi }}{{54}}\end{array} \right.\)

Mà \(\frac{\pi }{{18}} < \frac{{7\pi }}{{54}}\) suy ra giá trị nghiệm dương nhỏ nhất là \[{{\rm{x}}_0} = \frac{\pi }{{18}}\]

Vậy ta chọn đáp án B.