Answer
$$L = 12$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {{t^2},\frac{{4\sqrt 2 }}{3}{t^{3/2}},2t} \right\rangle ,\,\,\,\,\,for\,\,\,\,\,1 \leqslant t \leqslant 3 \cr
& {\text{Calculate }}{\bf{r}}'\left( t \right){\text{ and }}\left| {{\bf{r}}'\left( t \right)} \right| \cr
& {\bf{r}}'\left( t \right) = \left\langle {2t,\frac{{4\sqrt 2 }}{3}\left( {\frac{3}{2}} \right){t^{1/2}},2} \right\rangle \cr
& {\bf{r}}'\left( t \right) = \left\langle {2t,2\sqrt 2 {t^{1/2}},2} \right\rangle \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {{{\left( {2t} \right)}^2} + {{\left( {2\sqrt 2 {t^{1/2}}} \right)}^2} + {{\left( 2 \right)}^2}} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {4{t^2} + 8t + 4} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {4\left( {{t^2} + 2t + 1} \right)} \cr
& \left| {{\bf{r}}'\left( t \right)} \right| = 2\left( {t + 1} \right) \cr
& {\text{The Arc Length is }}\int_a^b {\left| {{\bf{r}}'\left( t \right)} \right|} dt \cr
& L = \int_1^3 {2\left( {t + 1} \right)} dt \cr
& {\text{Integrate}} \cr
& L = \left[ {{{\left( {t + 1} \right)}^2}} \right]_1^3 \cr
& L = {\left( {3 + 1} \right)^2} - {\left( {1 + 1} \right)^2} \cr
& L = 16 - 4 \cr
& L = 12 \cr} $$
