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