The rejection region for testing the null $H_0:\sigma^2=1$ against the alternative $H_1:\sigma^2>1$ is of the form $T>k$, where $T=\sum\limits_{k=1}^{10} X_i^2$ and $k$ is so chosen that size of the test is $0.05$.
You also have $$\frac T{\sigma^2}\sim \chi^2_{10}$$
This implies $$P_{H_0}\left[T>k\right]=P_{H_0}\left[\chi^2_{10}>k\right]=0.05\,,$$
so that in terms of the upper $5\%$ point of a chi-square distribution you have $$k=\chi^2_{10,0.05}\,,$$
whose value can be found from a chi-square table or software.
So the test function is
$$\varphi=\begin{cases}1&,\text{ if }T>\chi^2_{10,0.05} \\ 0 &,\text{ else }\end{cases}$$
Now the power at $\sigma^2=\sigma_1^2(>1)$ is
\begin{align}
E_{\sigma_1^2}[\varphi]&=P_{\sigma_1^2}\left[\frac T{\sigma_1^2}>\frac1{\sigma_1^2}\chi^2_{10,0.05}\right]
\\&=P\left[\chi^2_{10}>\frac1{\sigma_1^2}\chi^2_{10,0.05}\right]
\end{align}
This would give you the value of $\sigma_1^2$ for which this power equals $0.95$:
$$\frac1{\sigma_1^2}\chi^2_{10,0.05}=\chi^2_{10,0.95} \implies \sigma_1^2 = \frac{\chi^2_{10,0.05}}{\chi^2_{10,0.95}} = k' \,(\text{say})$$
The nature of the power function $E_{\sigma^2}[\varphi]=P_{\sigma^2}\left[T>\chi^2_{10,0.05}\right]$ (i.e. increasing/decreasing in $\sigma^2$) would then suggest the possible values of $\sigma^2_1$ for which $E_{\sigma_1^2}[\varphi]>0.95$.
Let $n$ be the sample size. Since $S^2$ is calculated from the sample generated by the $X_i$ we know that
$$X_i \sim N(\mu, \sigma^2)$$
$$\frac{(n-1)S^2}{\sigma^2}\sim \chi^2(n-1)$$
The distribution of the sum of a Gaussian rv and a Chi-Squared rv is an instance of the Generalized Chi-Squared Distribution. A variable $\xi$ with the Generalized Chi-Squared Distribution can be defined as follows:
$$\xi = x +\sum_1^n w_i y_i \text { where } x\sim N(m,s),\;\; w_i \in \mathbb{R},\;\;y_i \sim \chi'^2(k_i,\lambda_i)\text{ and } y_i \text { independent}$$
Note that $\chi'^2(k_i,\lambda_i)$ is the non-central Chi-squared distribution, it is related to the Chi-squared as follows:
$$\chi^2(n) = \chi'^2(n,0)$$
In your case, we want the sum of a single Chi-Squared variable and a Normal, where the Chi-Squared is based on the sample variance of the sample from the Normal random variable:
$$\xi = x + wy \;\;\text{ where } x\sim N\left(\mu,\frac{\sigma^2}{n}\right), \;y \sim \chi'^2(n-1,0)$$
What about the weight variable $w$? If $w=1$ then $\xi$ represents the following sum:
$$\bar{X} + \frac{(n-1)S^2}{\sigma^2}$$
But we want:
$$\bar{X} + S^2$$
Therefore, to get to this sum, we need to alter the weight applied to the chi-squared variable $y$:
$$w = \frac{\sigma^2}{n-1}$$
With this we have what we need:
$$\bar{X} + S^2 \sim \xi = x + wy \;\;\\\text{ where } x\sim N\left(\mu,\frac{\sigma^2}{n}\right), \;y \sim \chi'^2(n-1,0),\;w = \frac{\sigma^2}{n-1}$$
In summary
$$\bar{X} + S^2 \sim \tilde{\chi}^2\left(\frac{\sigma^2}{n-1},n-1,0,\mu,\sigma\right) $$
Best Answer
$\displaystyle \sum_{i=1}^N \left(\frac{x_i-\mu}{\sigma}\right)^2$ has a $\chi_N^2$ distribution (chi-squared with $N$ degrees of freedom) as as the sum of $N$ independent standard normal random variables.
So if $\mu=0$ and $\displaystyle s_2^2=\frac{1}{N}\sum_{i=1}^Nx_i^2$ then $N s_2^2 / \sigma^2$ also has a $\chi_N^2$ distribution.