The correct answer is $\boxed{\frac{{{{\text{e}}^4}}}{{4!}}}$.
The Taylor series expansion of $f(x)$ about $x=a$ is given by
$$f(x) = \sum_{n=0}^{\infty} \frac{{{f}^{(n)}}(a){{(x – a)}^n}}{{n!}}$$
where $f^{(n)}(x)$ is the $n$th derivative of $f(x)$.
In this case, $f(x) = {e^x}$ and $a=2$. Therefore, the Taylor series expansion of $e^x$ about $x=2$ is given by
$$e^x = \sum_{n=0}^{\infty} \frac{{e^{(n)}}(2){{(x – 2)}^n}}{{n!}}$$
The fourth derivative of $e^x$ is $e^x$. Therefore, the coefficient of $(x-2)^4$ is $\frac{{e^4}}{{4!}} = \boxed{\frac{{{{\text{e}}^4}}}{{4!}}}$.
The other options are incorrect because they do not correspond to the coefficient of $(x-2)^4$ in the Taylor series expansion of $e^x$ about $x=2$.