I am try to find the one-dimensional sufficient statistic for $\theta$. Although, I'm a bit confused about how you would arrive to the sufficient statistic.
Here is the common density function:
$f\left(x_{i} ; \theta\right)=\frac{1}{\theta \sqrt{\pi}} e^{\frac{-x_{i}^{2}}{\theta^{2}}}$, $-\infty<x_{i}<\infty, i=1,2, \ldots, n,$ where $\theta>0$ is unknown parameter.
Please let me know if you need me to clarify anything.
Yes, to use the factorization theorem, which says that if $f(x|\theta)=u(x)v[r(x),\theta]$, then r(x) is a sufficient statistic for theta.
$$f(x)=\frac{1}{\theta\sqrt \pi}e^{-\frac{x^2}{\theta^2}}$$
Then for a sample,
$$\begin{split}f(x)&=\prod\frac{1}{\theta\sqrt \pi}e^{-\frac{x^2}{\theta^2}}\\ &=\underbrace{1}_{u(x)}\cdot \underbrace{\frac{1}{\theta^n\pi^{n/2}}e^{-\frac{1}{\theta^2}\sum x^2}}_{=v[r(x),\theta]}\end{split}$$
Hence $\sum x^2$ is sufficient for theta.