$\triangle{DEF}$ is inscribed in circle $G$. In circle $G$, $\angle DFE = 60^\circ$. If the circumference of circle $G$ is $24$ units, what is the length of $\widehat{DE}$? Express your answer as a common fraction.
The measure of arc $\widehat{DE}$ is equal to the measure of $\angle DGE$. Since $\triangle DGE$ is an isosceles triangle, $\angle DGE = \angle DEG$, so $\angle DGE = \angle DEG = 60^\circ$ as well. The sum of the angle measures in quadrilateral $DAEB$ is $360^\circ$, so we have \begin{align*}
\angle DAB + \angle DEA + \angle DEG + \angle DGE &= 360^\circ
\\\Rightarrow\qquad \angle DAB + \angle DEA + 60^\circ + 60^\circ &= 360^\circ
\\\Rightarrow\qquad \angle DAB + \angle DEA &= 180^\circ
\end{align*} So $\triangle DAE$ is a triangle with an angle sum of $180^\circ$. Therefore, $DE$ is a straight line segment, so $\widehat{DE}$ is the entire circumference of circle $G$, or $\boxed{24}$ units.
We know that the circumference of a circle can be found using the formula $C = 2\pi r$, where $C$ is the circumference and $r$ is the radius. In this problem, we are given that the circumference of the circle is $24$ units.
Thus, $24 = 2\pi r$. To find the radius, we can solve for $r$:
$$r = \frac{24}{2\pi} = \frac{12}{\pi}$$
Since $\angle DFE = 60^\circ$, this angle subtends an arc on the circle with the same measure. Let's call this arc $\widehat{DE}$. We want to find the length of $\widehat{DE}$.
The circumference of a circle is $2\pi r$, and since the arc $\widehat{DE}$ is a portion of the circumference, the length of $\widehat{DE}$ can be found using the formula:
$$\widehat{DE} = \frac{\text{arc angle}}{360^\circ} \times \text{circumference}$$
In this case, the arc angle is $60^\circ$, and the circumference is $24$ units. Substituting these values into the formula, we get:
$$\widehat{DE} = \frac{60^\circ}{360^\circ} \times 24$$
Simplifying, we find:
$$\widehat{DE} = \frac{60}{360} \times 24 = \frac{1}{6} \times 24 = \boxed{4} \text{ units}$$
Therefore, the length of $\widehat{DE}$ is $4$ units.