Answer
$$\kappa \left( t \right) = \frac{{2\left| a \right|}}{{{{\left( {1 + 4{a^2}{t^2}} \right)}^{3/2}}}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {t,a{t^2}} \right\rangle \cr
& {\text{The parametrc curve is }}{\bf{r}}\left( t \right) = \left\langle {f\left( t \right),g\left( t \right)} \right\rangle \cr
& f\left( t \right) = t \cr
& g\left( t \right) = a{t^2} \cr
& {\text{Calculate the first and second derivatives of }}f\left( t \right){\text{ and }}g\left( t \right) \cr
& f'\left( t \right) = 1 \cr
& g'\left( t \right) = 2at \cr
& f''\left( t \right) = 0 \cr
& g''\left( t \right) = 2a \cr
& {\text{Use the formula }}\kappa \left( t \right) = \frac{{\left| {f'g'' - f''g'} \right|}}{{{{\left( {{{\left( {f'} \right)}^2} + {{\left( {g'} \right)}^2}} \right)}^{3/2}}}} \cr
& \kappa \left( t \right) = \frac{{\left| {\left( 1 \right)\left( {2a} \right) - \left( 0 \right)\left( {2at} \right)} \right|}}{{{{\left( {{{\left( 1 \right)}^2} + {{\left( {2at} \right)}^2}} \right)}^{3/2}}}} \cr
& \kappa \left( t \right) = \frac{{\left| {2a} \right|}}{{{{\left( {1 + 4{a^2}{t^2}} \right)}^{3/2}}}} \cr
& \kappa \left( t \right) = \frac{{2\left| a \right|}}{{{{\left( {1 + 4{a^2}{t^2}} \right)}^{3/2}}}} \cr} $$
