The correct answer is $\boxed{\frac{2}{3}}$.
The limit $\mathop {\lim }\limits_{{\text{x}} \to 0} \frac{{\sin \left[ {\frac{2}{3}{\text{x}}} \right]}}{{\text{x}}}$ can be evaluated using L’Hôpital’s rule. L’Hôpital’s rule states that if $\mathop {\lim }\limits_{{\text{x}} \to 0} \frac{{f\left( {\text{x}} \right)}}{{g\left( {\text{x}} \right)}}$ exists and $\mathop {\lim }\limits_{{\text{x}} \to 0} \frac{{f’\left( {\text{x}} \right)}}{{g’\left( {\text{x}} \right)}}$ also exists, then $\mathop {\lim }\limits_{{\text{x}} \to 0} \frac{{f\left( {\text{x}} \right)}}{{g\left( {\text{x}} \right)}} = \mathop {\lim }\limits_{{\text{x}} \to 0} \frac{{f’\left( {\text{x}} \right)}}{{g’\left( {\text{x}} \right)}}$.
In this case, we have $f\left( {\text{x}} \right) = \sin \left[ {\frac{2}{3}{\text{x}}} \right]$ and $g\left( {\text{x}} \right) = {\text{x}}$. Therefore, we have
$$\begin{align}
\mathop {\lim }\limits_{{\text{x}} \to 0} \frac{{\sin \left[ {\frac{2}{3}{\text{x}}} \right]}}{{\text{x}}} &= \mathop {\lim }\limits_{{\text{x}} \to 0} \frac{{f’\left( {\text{x}} \right)}}{{g’\left( {\text{x}} \right)}} \\
&= \mathop {\lim }\limits_{{\text{x}} \to 0} \frac{{2\cos \left[ {\frac{2}{3}{\text{x}}} \right] \cdot \frac{2}{3}}}{{1}} \\
&= \frac{2}{3}.
\end{align}$$
Therefore, the limit $\mathop {\lim }\limits_{{\text{x}} \to 0} \frac{{\sin \left[ {\frac{2}{3}{\text{x}}} \right]}}{{\text{x}}}$ is $\boxed{\frac{2}{3}}$.
The other options are incorrect because they do not represent the correct value of the limit.