In response to:

So how do I find $E(x^2_i)$?

Assuming there are no issues with the steps in the derivation up to that point, you can use the following standard result relating expectation, variance and 2nd moments

$$\mathbb{E}[X^2_i] = \text{Var}(X_i) + \mathbb{E}[X_i]^2$$

I know nothing of the Rayleigh distribution, but according to wikipedia, it has mean

$$\mathbb{E}[X_i] = \sigma \sqrt{\frac{\pi}{2}}$$

and variance

$$\text{Var}(X_i) = \frac{4 - \pi}{2} \sigma^2$$

Using these facts we have that the expectation of the estimator $\hat{\sigma}^2$ is

\begin{align*} \mathbb{E}[\hat{\sigma^2}] &= \frac{1}{2n} \sum_{i=1}^n \mathbb{E}[X_i^2] \\ &= \frac{1}{2n} \sum_{i=1}^n \left( \frac{4 - \pi}{2} \sigma^2 + \frac{\pi}{2} \sigma^2 \right) \\ &= \frac{2n \sigma^2}{2n} \\ &= \sigma^2 \end{align*}

Which is unbiased as $\mathbb{E}[\hat{\sigma}^2] = \sigma^2$.

In response to:

So how do I find $E(x^2_i)$?

Assuming there are no issues with the steps in the derivation up to that point, you can use the following standard result relating expectation, variance and 2nd moments

$$\mathbb{E}[X^2_i] = \text{Var}(X_i) + \mathbb{E}[X_i]^2$$

I know nothing of the Rayleigh distribution, but according to wikipedia, it has mean

$$\mathbb{E}[X_i] = \sigma \sqrt{\frac{\pi}{2}}$$

and variance

$$\text{Var}(X_i) = \frac{4 - \pi}{2} \sigma^2$$

Using these facts we have that the expectation of the estimator $\hat{\sigma}^2$ is

\begin{align*} \mathbb{E}[\hat{\sigma}^2] &= \frac{1}{2n} \sum_{i=1}^n \mathbb{E}[X_i^2] \\ &= \frac{1}{2n} \sum_{i=1}^n \left( \frac{4 - \pi}{2} \sigma^2 + \frac{\pi}{2} \sigma^2 \right) \\ &= \frac{2n \sigma^2}{2n} \\ &= \sigma^2 \end{align*}

Therefore the estimator $\hat{\sigma}^2(X_1, ..., X_n)$ is unbiased as $\mathbb{E}[\hat{\sigma}^2] = \sigma^2$.