Answer
$$K = \frac{{12}}{{145\sqrt {145} }},{\text{ }}r = \frac{{145\sqrt {145} }}{{12}}$$
Work Step by Step
$$\eqalign{
& y = {x^3},{\text{ }}x = 2 \cr
& {\text{Calculate the curvature, use }}K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}} \cr
& {\text{Find }}y'\left( x \right){\text{ and }}y''\left( x \right),{\text{ }} \cr
& y'\left( x \right) = \frac{d}{{dx}}\left[ {{x^3}} \right] \cr
& y'\left( x \right) = 3{x^2} \cr
& {\text{Evaluate at }}x = 2 \cr
& y'\left( 2 \right) = 3{\left( 2 \right)^2} = 12 \cr
& y''\left( x \right) = \frac{d}{{dx}}\left[ {3{x^2}} \right] \cr
& y''\left( x \right) = 6x \cr
& {\text{Evaluate at }}x = 2 \cr
& y''\left( 2 \right) = 12 \cr
& \underbrace {K = \frac{{\left| {y''\left( x \right)} \right|}}{{{{\left( {1 + {{\left[ {y'\left( x \right)} \right]}^2}} \right)}^{3/2}}}}}_ \Downarrow \cr
& {\text{ at }}x = 2 \cr
& K = \frac{{\left| {12} \right|}}{{{{\left( {1 + {{\left[ {12} \right]}^2}} \right)}^{3/2}}}} \cr
& K = \frac{{12}}{{145\sqrt {145} }} \cr
& {\text{The radius of curvature is }}r = \frac{1}{K} \cr
& r = \frac{{145\sqrt {145} }}{{12}} \cr} $$
