Answer
$\pi-3$
Work Step by Step
As we are given that $r=8 \sin^3 (\dfrac{\theta}{3})$
This gives : $r'=8 \sin^2 (\dfrac{\theta}{3}) \cos (\dfrac{\theta}{3})$
The length of the curve: $L= \int_{0}^{\pi/4} \sqrt{r^2+r'^2} d\theta$
$\implies L=64 \int_{0}^{\pi/4} \sqrt {[\sin^{4} (\dfrac{\theta}{3})} d\theta$
This implies that
$L=[4 (\theta)-(6) \sin(\dfrac{2\theta}{3})]_{0}^{(\pi/4)} =\pi-3$
