Answer
$$c = \root 3 \of {\frac{{65}}{{\ln 64}}} $$
Work Step by Step
$$\eqalign{
& f\left( x \right) = \ln x,{\text{ }}g\left( x \right) = {x^3}{\text{ }}\underbrace {\left[ {1,4} \right]}_{\left[ {a,b} \right]} \cr
& {\text{Differentiate both functions}} \cr
& f'\left( x \right) = \frac{d}{{dx}}\left[ {\ln x} \right] = \frac{1}{x} \cr
& g'\left( x \right) = \frac{d}{{dx}}\left[ {{x^3}} \right] = 3{x^2} \cr
& {\text{Apply the Extended Mean Value Theorem}} \cr
& \frac{{f'\left( c \right)}}{{g'\left( c \right)}} = \frac{{f\left( b \right) - f\left( a \right)}}{{g\left( b \right) - g\left( a \right)}} \cr
& {\text{,then}} \cr
& \frac{{\frac{1}{c}}}{{3{c^2}}} = \frac{{\ln \left( 4 \right) - \ln \left( 1 \right)}}{{{{\left( 4 \right)}^3} - {{\left( 1 \right)}^3}}} \cr
& {\text{Simplifying}} \cr
& \frac{1}{{3{c^3}}} = \frac{{\ln 4}}{{65}} \cr
& {c^3} = \frac{{65}}{{3\ln 4}} \cr
& c = \root 3 \of {\frac{{65}}{{\ln 64}}} \cr} $$
