Due to independence, joint density of the sample $\mathbf X=(X_1,X_2,\cdots,X_n)$ is
\begin{align}
f_{\theta}(\mathbf x)&=\prod_{i=1}^n \frac{\beta}{\theta^{\beta}}x_i^{\beta -1}e^{-x_i^{\beta}/{\theta}^{\beta}}\mathbf1_{x_i>0}
\\&=\frac{e^{-\frac{\sum x_i^{\beta}}{\theta^{\beta}}}}{\theta^{n\beta}}\beta^n \left(\prod_{i=1}^n x_i\right)^{\beta-1}\mathbf1_{x_1,\cdots,x_n>0}\quad,\theta,\beta>0
\\&=g(\theta,t(\mathbf x))h(\mathbf x)
\end{align}
, where $g(\theta, t(\mathbf x))= \frac{e^{-\frac{\sum x_i^{\beta}}{\theta^{\beta}}}}{\theta^{n\beta}} $ depends on $\theta$ and on $x_1,x_2,\cdots,x_n$ through $t(\mathbf x)=\sum_{i=1}^n x_i^{\beta}$ and $h(\mathbf x)= \beta^n \left(\prod_{i=1}^n x_i\right)^{\beta-1} $ is independent of $\theta$.
Assuming $\beta$ is known, by the Factorization theorem, a sufficient statistic for $\theta$ would be
$$T(\mathbf X)=\sum_{i=1}^nX_i^{\beta}$$
Your answer is not quite right. But we can say that $e^{-T}$ is also sufficient for $\theta$, being a bijective function of $T$.
Best Answer
Hint:
$$f(x_1,\dots,x_n|\sigma^2)=\left(\frac{1}{\sqrt{2\pi\sigma^2}}\right)^ne^{-\frac{1}{2\sigma^2}\sum_{i=1}^n(x_i-\mu)^2}$$
So if you let
$$T(X_1,\dots,X_n)=\sum_{i=1}^n(x_i-\mu)^2$$
could you use the factorization theorem to conclude $T$ is sufficient?