Your strategy is a good one. Let's justify it.
In a sample $x_1, x_2, \ldots, x_m$ of $m$ independent draws of $X,$ the statistic
$$(x_1-\mu_1)^2 + \cdots + (x_m-\mu_1)^2$$
behaves like $\sigma_1^2$ times a $\chi^2(m)$ distribution. A similar result holds for an independent sample of $n$ draws from $Y.$ Therefore, under the null hypothesis $\sigma_1^2=10\sigma_2^2,$ the ratio
$$F=\frac{10\left[x_1-\mu_1)^2 + \cdots + (x_m-\mu_1)^2\right]/m}{\left[(y_1-\mu_2)^2 + \cdots + (y_n-\mu_2)^2\right]/n}$$
behaves like $10\sigma_1^2 / (10\sigma_1^2) = 1$ times an $F(m,n)$ distribution. For a symmetric two-tailed test with confidence $1-\alpha,$ compare this ratio to the $\alpha/2$ and $1-\alpha/2$ quantiles of this $F$ distribution.
The principal noteworthy element of this result is that the parameters of $F$ (its "degrees of freedom") are $m$ and $n$ rather than the conventional $m-1$ and $n-1$ used for distributions with unknown variances.
For example, here is a histogram of $F$ for 100,000 simulated datasets with $m=3$ and $n=5$ where the null hypothesis holds. (Because values of $F$ can be extremely large for small sample sizes, I have drawn it on a logarithmic scale.) On it is superimposed in red the density of the $F(3,5)$ distribution: it's a close fit, supporting the accuracy of the previous calculations and confirming that $(m,n)$ are the correct values for the degrees of freedom.

The vertical dotted lines are the $0.025$ and $1-0.025$ limits for a test of $100(1-0.05)=95\%$ confidence. These limits are approximately $0.067$ and $7.76.$ Thus, with these sample sizes of $m$ and $n,$ you would reject the null when the sample ratio $F$ is less than $0.067$ or greater than $7.76.$