Answer
$$s = \frac{{80\sqrt {10} - 8}}{{27}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = {\bf{i}} + {t^2}{\bf{j}} + {t^3}{\bf{k}},{\text{ }}\left[ {0,2} \right] \cr
& {\text{Differentiate }}{\bf{r}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {{\bf{i}} + {t^2}{\bf{j}} + {t^3}{\bf{k}}} \right] \cr
& {\bf{r}}'\left( t \right) = 2t{\bf{j}} + 3{t^2}{\bf{k}} \cr
& {\text{Find }}\left\| {{\bf{r}}'\left( t \right)} \right\| \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{{\left( 0 \right)}^2} + {{\left( {2t} \right)}^2} + {{\left( {3{t^2}} \right)}^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {4{t^2} + 9{t^4}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = t\sqrt {4 + 9{t^2}} \cr
& {\text{Find the arc length }}s{\text{ using the formula }}s = \int_a^b {\left\| {{\bf{r}}'\left( t \right)} \right\|} dt \cr
& s = \int_0^2 {t\sqrt {4 + 9{t^2}} } dt \cr
& s = \frac{1}{{18}}\int_0^2 {18t\sqrt {4 + 9{t^2}} } dt \cr
& {\text{Integrate}} \cr
& s = \frac{1}{{18}}\left[ {\frac{{{{\left( {4 + 9{t^2}} \right)}^{3/2}}}}{{3/2}}} \right]_0^2 \cr
& s = \frac{1}{{27}}\left[ {{{\left( {4 + 9{t^2}} \right)}^{3/2}}} \right]_0^2 \cr
& s = \frac{1}{{27}}\left[ {{{\left( {4 + 9{{\left( 2 \right)}^2}} \right)}^{3/2}} - {{\left( {4 + 9{{\left( 0 \right)}^2}} \right)}^{3/2}}} \right] \cr
& s = \frac{1}{{27}}\left[ {{{\left( {40} \right)}^{3/2}} - {{\left( 4 \right)}^{3/2}}} \right] \cr
& s = \frac{{40\sqrt {40} - 8}}{{27}} \cr
& s = \frac{{80\sqrt {10} - 8}}{{27}} \cr
& {\text{Graph}} \cr} $$
