Answer
$c=\dfrac{1+\sqrt 7}{3}, \dfrac{1-\sqrt 7}{3}$
Work Step by Step
Since, the function $f(x)=x^3-x^2$ is continuous on $[-1,2]$ and differentiable on $(-1,2)$ and $f'(x)=3x^2-2x$ .
The Mean value Theorem states that there is a point $c$ and $c \in (-1,2)$
such that $f'(c)=\dfrac{f(2)-f(-1)}{[2-(-1)]}=\dfrac{6}{3}=2$
Now,we have $f'(x)=3x^2-2x$
so, $f'(c)=3c^2-2c=2$ or, $3c^2-2c-2=0$
Thus, $c=\dfrac{1+\sqrt 7}{3}, \dfrac{1-\sqrt 7}{3}$
