Answer
$$
\text { The length }
=\int_{0}^{2\pi} \sin^2\theta\sqrt{1 +8\cos^2\theta} d \theta.
$$
Work Step by Step
Since $r=f(\theta)= \sin^3\theta$, then $f'(\theta)=3\sin^2\theta\cos\theta$
The length is given by
$$
\text { The length }=\int_{0}^{2\pi} \sqrt{f(\theta)^{2}+f^{\prime}(\theta)^{2}} d \theta\\
=\int_{0}^{2\pi} \sqrt{\sin^6\theta +9\sin^4\theta\cos^2\theta} d \theta\\
=\int_{0}^{2\pi} \sin^2\theta\sqrt{\sin^2\theta +9\cos^2\theta} d \theta\\
=\int_{0}^{2\pi} \sin^2\theta\sqrt{1 +8\cos^2\theta} d \theta.
$$
