The correct answer is $\boxed{\frac{1}{2}}$.
We can use the squeeze theorem to solve this problem. The squeeze theorem states that if $f(x) \leq g(x) \leq h(x)$ for all $x$ in a given interval, and $\lim_{x \to a} f(x) = \lim_{x \to a} h(x)$, then $\lim_{x \to a} g(x)$ also exists and is equal to the same value.
In this case, we have $2 – \frac{{{{\text{x}}^2}}}{3} < \frac{{{\text{x}}\sin {\text{x}}}}{{1 – \cos {\text{x}}}} < 2$ for all $x$ close to 0. We also know that $\lim_{x \to 0} 2 – \frac{{{{\text{x}}^2}}}{3} = 1$ and $\lim_{x \to 0} 2 = 2$. Therefore, by the squeeze theorem, we have $\lim_{x \to 0} \frac{{{\text{x}}\sin {\text{x}}}}{{1 – \cos {\text{x}}}} = \frac{1}{2}$.
We can also solve this problem by using L’Hôpital’s rule. L’Hôpital’s rule states that if $\lim_{x \to a} \frac{f(x)}{g(x)}$ exists and $\lim_{x \to a} \frac{f'(x)}{g'(x)}$ also exists, then $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$.
In this case, we have $\lim_{x \to 0} \frac{{{\text{x}}\sin {\text{x}}}}{{1 – \cos {\text{x}}}} = \lim_{x \to 0} \frac{{\text{x}} \cos {\text{x}} + \sin {\text{x}}}{-\sin {\text{x}}} = \lim_{x \to 0} \frac{{\text{x}} \cos {\text{x}} + \sin {\text{x}}}{-\sin {\text{x}}} = \frac{1}{2}$.
Therefore, the value of $\lim_{x \to 0} \frac{{{\text{x}}\sin {\text{x}}}}{{1 – \cos {\text{x}}}}$ is $\boxed{\frac{1}{2}}$.