Answer
$c$ = $\frac{\sqrt 3}{9}$
Work Step by Step
The Mean Value Theorem states that if $f$ is a continuous function over the closed interval $[a,b]$ and differentiable over the open interval $(a,b)$, then there exists at least one point $c\in(a,b)$ so that
$$f′(c)=\dfrac{f(b)−f(a)}{b−a}.$$
The function $f(x)$ = $\sqrt[3]x$ is continuous on $[0,1]$ and differentiable on $(0,1)$.
According the the Mean Value Theorem, there is at least one point $c\in(0,1)$ so that we have:
$f'(c)=\dfrac{f(1)-f(0)}{1-0}$
$f'(c)=\dfrac{\sqrt[3]1-\sqrt[3]0}{1}$
$f'(c)=1$
But $f'(x)=\frac{1}{3}x^{-2/3}$, so we have:
$\dfrac{1}{3}c^{-2/3}=1$
$c^2=\frac{1}{27}$
$c=\pm\dfrac{1}{3\sqrt 3}=\pm\dfrac{\sqrt 3}{9}$