Answer
$$s = 6a$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = a{\cos ^3}t{\bf{i}} + a{\sin ^3}t{\bf{j}},{\text{ }}\left[ {0,2\pi } \right] \cr
& {\text{Differentiate }}{\bf{r}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {a{{\cos }^3}t{\bf{i}} + a{{\sin }^3}t{\bf{j}}} \right] \cr
& {\bf{r}}'\left( t \right) = - 3a{\cos ^2}t\sin t{\bf{i}} + 3a{\sin ^2}t\cos t{\bf{j}} \cr
& {\text{Find }}\left\| {{\bf{r}}'\left( t \right)} \right\| \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{{\left( { - 3a{{\cos }^2}t\sin t} \right)}^2} + {{\left( {3a{{\sin }^2}t\cos t} \right)}^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {9{a^2}{{\cos }^4}t{{\sin }^2}t + 9{a^2}{{\sin }^4}t{{\cos }^2}t} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {9{a^2}{{\cos }^2}t{{\sin }^2}t\left( {{{\cos }^2}t + {{\sin }^2}t} \right)} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = 3a\sin t\cos t \cr
& {\text{Find the arc length }}s{\text{ using the formula }}s = \int_a^b {\left\| {{\bf{r}}'\left( t \right)} \right\|} dt \cr
& {\text{From the graph we can set the arc length from }}\left[ {0,\frac{\pi }{2}} \right]{\text{ as}} \cr
& s = \int_0^{\pi /2} {3a\sin t\cos t} dt \cr
& {\text{Applying for }}\left[ {0,\frac{\pi }{2}} \right] \cr
& s = 4\int_0^{\pi /2} {3a\sin t\cos t} dt \cr
& s = 12\int_0^{\pi /2} {a\sin t\cos t} dt \cr
& {\text{Integrating}} \cr
& s = 12a\left[ {\frac{{{{\sin }^2}t}}{2}} \right]_0^{\pi /2} \cr
& s = 12a\left[ {\frac{{{{\sin }^2}\left( {\pi /2} \right)}}{2} - \frac{{{{\sin }^2}\left( 0 \right)}}{2}} \right] \cr
& s = 12a\left[ {\frac{1}{2} - \frac{0}{2}} \right] \cr
& s = 6a \cr
& {\text{Graph}} \cr} $$
