I am not sure how to do this. To find the method of moments estimator I did:
$E[X] = \frac{-\theta + \theta}{2} = 0$
use 2nd moment:
$E[X^2] = \frac{(-\theta)^2 + -(\theta^2) + \theta^2}{3} = \frac{\theta^2}{3} = \frac{1}{n} \sum_{i=1}^n X_i^2$
$\hat{\theta} = \sqrt{\frac{3}{n} \sum_{i=1}^n X_i^2}$
And to find if it is unbiased, I tried this, but am not sure what to do with the integral
$E[\hat{\theta}] = \sqrt{\frac{3}{n}} E[\sqrt{\sum_{i=1}^n X_i^2}]$
$E[\hat{\theta}] = \int_{-\theta}^\theta \sqrt{\frac{3}{n} \sum_{i=1}^n X_i^2} \frac{1}{2\theta} dx$
$= \sqrt{\frac{3}{n}} * \frac{1}{2\theta} \int_{-\theta}^\theta \sqrt{\sum_{i=1}^n X_i^2} dx$
It is sufficient to compute the expectation of $$\tilde \theta = \sqrt{\frac{3}{n} \sum_{i=1}^n X_i^2}$$ for the case $n = 1$ to show that it is biased. We have $$\operatorname{E}[\tilde \theta] = \int_{x=-\theta}^\theta \sqrt{3x^2} \cdot \frac{1}{2\theta} \, dx = \frac{\sqrt{3}}{\theta} \int_{x=0}^\theta x \, dx = \frac{\sqrt{3}}{2} \theta \ne \theta.$$ Moreover, because of Jensen's inequality applied to the concave function $\sqrt{x}$, we have $$\operatorname{E}[\tilde \theta] \le \sqrt{\operatorname{E}\left[\frac{3}{n} \sum_{i=1}^n X_i^2\right]} = \sqrt{3 \operatorname{E}[X^2]} = \theta$$ for all $n$. Equality is not attained because $\sqrt{x}$ is strictly concave for all positive reals.