Answer
$$\kappa \left( t \right) = \frac{{\sqrt {{t^4} + 4{t^2} + 1} }}{{{{\left( {1 + {t^2} + {t^4}} \right)}^{3/2}}}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = t{\bf{i}} + \frac{1}{2}{t^2}{\bf{j}} + \frac{1}{3}{t^3}{\bf{k}} \cr
& {\text{Calculate the derivatives }}{\bf{r}}'\left( t \right){\text{ and }}{\bf{r}}{\text{''}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ t \right]{\bf{i}} + \frac{d}{{dt}}\left[ {\frac{1}{2}{t^2}} \right]{\bf{j}} + \frac{d}{{dt}}\left[ {\frac{1}{3}{t^3}} \right]{\bf{k}} \cr
& {\bf{r}}'\left( t \right) = \left( 1 \right){\bf{i}} + t{\bf{j}} + {t^2}{\bf{k}} \cr
& and \cr
& {\bf{r}}''\left( t \right) = \frac{d}{{dt}}\left[ 1 \right]{\bf{i}} + \frac{d}{{dt}}\left[ t \right]{\bf{j}} + \frac{d}{{dt}}\left[ {{t^2}} \right]{\bf{k}} \cr
& {\bf{r}}''\left( t \right) = {\bf{j}} + 2t{\bf{k}} \cr
& \cr
& {\text{Calculate the cross product }}{\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right) \cr} $$
\[\begin{gathered}
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
1&t&{{t^2}} \\
0&1&{2t}
\end{array}} \right| \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left| {\begin{array}{*{20}{c}}
t&{{t^2}} \\
1&{2t}
\end{array}} \right|{\mathbf{i}} - \left| {\begin{array}{*{20}{c}}
1&{{t^2}} \\
0&{2t}
\end{array}} \right|{\mathbf{j}} + \left| {\begin{array}{*{20}{c}}
1&t \\
0&1
\end{array}} \right|{\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = \left( {2{t^2} - {t^2}} \right){\mathbf{i}} - \left( {2t - 0} \right){\mathbf{j}} + \left( {1 - 0} \right){\mathbf{k}} \hfill \\
{\mathbf{r}}'\left( t \right) \times {\mathbf{r}}''\left( t \right) = {t^2}{\mathbf{i}} - 2t{\mathbf{j}} + {\mathbf{k}} \hfill \\
\end{gathered} \]
$$\eqalign{
& {\text{Use the formula }}\left( 3 \right)\,\,\,\kappa \left( t \right) = \frac{{\left\| {{\bf{r}}'\left( t \right) \times {\bf{r}}''\left( t \right)} \right\|}}{{{{\left\| {{\bf{r}}'\left( t \right)} \right\|}^3}}}.{\text{ Thus}}{\text{,}} \cr
& \kappa \left( t \right) = \frac{{\left\| {{t^2}{\bf{i}} - 2t{\bf{j}} + {\bf{k}}} \right\|}}{{{{\left\| {{\bf{i}} + t{\bf{j}} + {t^2}{\bf{k}}} \right\|}^3}}} \cr
& \kappa \left( t \right) = \frac{{\sqrt {{t^4} + 4{t^2} + 1} }}{{{{\left( {\sqrt {1 + {t^2} + {t^4}} } \right)}^3}}} \cr
& \kappa \left( t \right) = \frac{{\sqrt {{t^4} + 4{t^2} + 1} }}{{{{\left( {1 + {t^2} + {t^4}} \right)}^{3/2}}}} \cr} $$
