Answer
$\dfrac{\pi}{6}$
Work Step by Step
We know that the trigonometric function $\sec^{-1} x$ has range: $[0, \pi]$.
Here, $\sec^{-1} (\dfrac{2 \sqrt 3}{3})=\dfrac{\pi}{6}$, because $\sec(\dfrac{\pi}{6})=\dfrac{2 \sqrt 3}{3}$ and $\dfrac{\pi}{6}$ lies in the range of $\sec^{-1}{x}$ .