Answer
$$\frac{4\sqrt 3}{9}.$$
Work Step by Step
Since $(\tan\left(3 t-\frac{\pi}{6}\right) )'=3\sec^{2}\left(3 t-\frac{\pi}{6}\right) $, we have
$$\int_{0}^{ \pi/6 }\sec ^{2}\left(3 t-\frac{\pi}{6}\right) d t= \frac{1}{3}\tan\left(3 t-\frac{\pi}{6}\right)|_{0 }^{ \pi/6 }\\
=\frac{1}{3} \tan\left(\frac{\pi}{2}-\frac{\pi}{6}\right)+\frac{1}{3}\tan\left(\frac{\pi}{6}\right)\\
=\frac{1}{3}\tan\left(\frac{\pi}{3}\right)+\frac{1}{3}\tan\left(\frac{\pi}{6}\right)=\frac{1}{3}[\sqrt 3+\frac{\sqrt 3}{3}]\\
=\frac{4\sqrt 3}{9}.$$
