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.
