Let the first term be $a$ and the common ratio be $r$. Therefore,
$$ar=2\ \ (1) \qquad \text{and} \qquad ar^3=6\ \ (2)$$
Dividing $(2)$ by $(1)$ eliminates the $a$, yielding $r^2=3$, so $r=\pm\sqrt{3}$.
Now, since $ar=2$, $a=\frac{2}{r}$, so $a=\frac{2}{\pm\sqrt{3}}=\pm\frac{2\sqrt{3}}{3}$.
We therefore see that $\boxed{\textbf{(B)}\ -\frac{2\sqrt{3}}{3}}$ is a possible first term.