Answer
$$\left( {4\sqrt 2 ,\frac{{2\pi }}{3},\frac{\pi }{4}} \right)$$
Work Step by Step
$$\eqalign{
& \left( { - 2,2\sqrt 3 ,4} \right) \cr
& \left( {x,y,z} \right):{\text{ }}\left( { - 2,2\sqrt 3 ,4} \right) \to x = - 2,{\text{ }}y = 2\sqrt 3 ,{\text{ }}z = 4 \cr
& {\text{Rectangular to spherical }}\left( {\rho ,\theta ,\phi } \right) \cr
& {\rho ^2} = {x^2} + {y^2} + {z^2} \to \rho = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( {2\sqrt 3 } \right)}^2} + {{\left( 4 \right)}^2}} = 4\sqrt 2 \cr
& \tan \theta = \frac{y}{x} \to \theta = {\tan ^{ - 1}}\left( {\frac{{2\sqrt 3 }}{{ - 2}}} \right) = - \frac{\pi }{3} + \pi = \frac{{2\pi }}{3} \cr
& \phi = \arccos \left( {\frac{z}{{\sqrt {{x^2} + {y^2} + {z^2}} }}} \right) \to \phi = \arccos \left( {\frac{4}{{4\sqrt 2 }}} \right) = \frac{\pi }{4} \cr
& {\text{The spherical }}\left( {\rho ,\theta ,\phi } \right){\text{ coordinates are:}} \cr
& \left( {4\sqrt 2 ,\frac{{2\pi }}{3},\frac{\pi }{4}} \right) \cr} $$
