Answer
$$L=\frac{8}{3}\left[\left(\pi^{2}+1\right)^{3 / 2}-1\right]$$
Work Step by Step
Given
$$
r=\theta^{2}, \quad 0 \leqslant \theta \leqslant 2 \pi
$$
The length is given by
\begin{align*}
L&=\int_{a}^{b} \sqrt{r^{2}+\left(\frac{d r}{ d \theta}\right)^{2}} d \theta\\
&=\int_{0}^{2 \pi} \sqrt{\left(\theta^{2}\right)^{2}+(2 \theta)^{2}} d \theta\\
&=\int_{0}^{2 \pi} \sqrt{\theta^{4}+4 \theta^{2}} d \theta\\
&=\int_{0}^{2 \pi} \theta \sqrt{\theta^{2}+4} d \theta\\
&=\frac{1}{2}\frac{2}{3}(\theta^{2}+4)^{3/2}\bigg|_{0}^{2\pi}\\
&=\frac{1}{3}\left[4^{3 / 2}\left(\pi^{2}+1\right)^{3 / 2}-4^{3 / 2}\right]\\
&=\frac{8}{3}\left[\left(\pi^{2}+1\right)^{3 / 2}-1\right]
\end{align*}
