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Taking $L$ to be the Hilbert class field of $K$, such a construction would imply that the Steinitz class of $L/K$ is always trivial. Yet this is false - take $K = \mathbb{Q}(\sqrt{-15})$ for example.

(EDIT): If $L = K(\sqrt{\alpha})$ is a tamely ramified extension of $K$, then the Steinitz class is represented by an ideal $I$ such that $I^2 (\alpha) = \Delta_{L/K}$. In particular, if $L/K$ is unramified, so $(\alpha) = \mathfrak{n}^2$, then the Steinitz class is trivial if and only if $\mathfrak{n}$ is principal, i.e., if and only if we may take $\alpha$ to be a unit in $K$. Clearly this is not the case for $L/K$ above.

In fact, I just found a source that works out this example in explicit detail - see Theorems 2.2 and 3.1 of the following:

Taking $L$ to be the Hilbert class field of $K$, such a construction would imply that the Steinitz class of $L/K$ is always trivial. Yet this is false - take $K = \mathbb{Q}(\sqrt{-15})$ for example.