Answer
$$\kappa \left( x \right) = \frac{2}{{{{\left( {1 + 4{x^2}} \right)}^{3/2}}}}$$
Work Step by Step
$$\eqalign{
& f\left( x \right) = {x^2} \cr
& {\text{Calculate }}f'\left( x \right){\text{ and }}f''\left( x \right) \cr
& f'\left( x \right) = 2x \cr
& f''\left( x \right) = 2 \cr
& {\text{Use the formula }}\kappa \left( x \right) = \frac{{\left| {f''\left( x \right)} \right|}}{{{{\left( {1 + f'{{\left( x \right)}^2}} \right)}^{3/2}}}} \cr
& \kappa \left( x \right) = \frac{{\left| 2 \right|}}{{{{\left( {1 + {{\left( {2x} \right)}^2}} \right)}^{3/2}}}} \cr
& {\text{Simplifying}} \cr
& \kappa \left( x \right) = \frac{2}{{{{\left( {1 + 4{x^2}} \right)}^{3/2}}}} \cr} $$
