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Edit: As requested, here are some more details, as well as a low-tech way of seeing why $L=K(\sqrt{-3})$ is the Hilbert class field of $K$.

First off, an integral basis for $L$ is $\{1, (1+\sqrt{-3})/2, \sqrt2, (\sqrt 2 + \sqrt{-6})/2\}$ (taken from an exercise in Marcus's "Number Fields"), and then a simple computation gives $\text{disc } \mathcal O_L = 12^2$. On the other hand, we find that $\text{disc } M = 1920^2$ and so the equation $\text{disc } M = (\mathcal O_L : M)^2 \text{disc } \mathcal O_L$ gives $(\mathcal O_L : M) = 160$.

Now, as to why $L$ is the Hilbert class field of $K$---well, the HCF has to be a quadratic extension of $K$ (because the class number of $K$ is easily computed to be $2$), so it suffices to show that the finite primes of $K$ are unramified in $L$. For this we can use the relative discriminant of $L/K$: this is an ideal that contains, in particular, $\text{disc } \{1,\sqrt 2\} = 8$ and $\text{disc } \{1, (1+\sqrt{-3})/2\} = -3$, hence contains $1$. That is, the relative discriminant of $L/K$ is the unit ideal and so the extension is unramified.

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How did you determine that the index $(\mathcal O_L : M)$ is $8$? It seems to me that it's actually $160$, which is divisible by $5$.

Incidentally, a quick way to see that Milne is correct is to note that the discriminant of $K$ is $-4\cdot 6$ and $6$ is an idoneal number: this means that the Hilbert class field of $K$ coincides with its genus field, which is easily computed to be $K(\sqrt{-3})$. See section 6 of Cox's "Primes of the form $x^2 + ny^2$".