Given a number field $K$, when is its Hilbert class field an abelian extension of $\mathbb{Q}$? I am going to be on the road soon, so pleas don't be offended if I don't respond quickly to a comment.

The genus class field of an extension $K/F$ is defined to be the largest extension $L/K$ with the following properties:
Thus the quick answer to your question is: the Hilbert class field of $K$ is abelian over ${\mathbb Q}$ if and only if the Hilbert class field of $K$ coincides with its genus class field. The notsoquick answer would tell you more about the construction of the genus class field. For abelian extensions of the rationals, the construction is easy: everything you'd like to know should be contained in Frölich's book
Basically you will have to look for the largest abelian extension of ${\mathbb Q}$ with the same conductor as $K/{\mathbb Q}$. 


Obivously, if the Hilbert class field $H$ of $K$ is abelian over $\mathbb Q$, then $K$ (which is a subfield of $H$) must be abelian over $\mathbb Q$. So suppose that this is the case. In general, there is then a maximal subfield of $H$ that is abelian over $\mathbb Q$; call it $F$. It certainly contains $K$, and is called the genus field of $K$. By class field theory $Gal(F/K)$ is a quotient of the class group $Cl(K)$ of $K$. Which quotient? At least when $Gal(K/\mathbb Q)$ is cyclic, it is the maximal quotient on which $Gal(K/\mathbb Q)$ acts trivially. If $K$ is quadratic then $Gal(K/\mathbb Q)$ acts on $Cl(K)$ by inversion, so $F/K$ corresponds to the maximal 2elementary abelian quotient of $Cl(K)$. In particular, in this case $F = H$ if and only if $Cl(K)$ is an elementary abelian 2group. (E.g. $\mathbb Q(\sqrt{5})$, whose class group is of order 2.) 


Is there a good/intuitive way to generate a stock of examples of nonabelian unramified extensions? The only examples I know of are a bit unintuitive (eg in Janusz's book). Intuition might suggest to start looking at Galois extensions with group equal to semidirect products. I was originally interested in this question by the necessity of "ab" on the rhs of $H^{1} (X_{Zar}\,, \mathcal{O}_{X} ^ {*}) = \pi_{1} ^{ab} (X), \; \; X = Spec\; \mathcal{O}_K$ ` (reinterpretation of unramified global cft). 


I happened to come across this question again today. In some cases at least, the Hilbert class field $H$ of an abelian extension $K$ of $\mathbf{Q}$ will have to be abelian over $\mathbf{Q}$ for purely algebraic reasons. Let $F$ be any field, $KF$ an abelian extension of group $G=\mathrm{Gal}(KF)$ and containing a primitive $n$th root of unity for some $n>1$, $\omega:G\to(\mathbf{Z}/n\mathbf{Z})^\times$ the cyclotomic character giving the action of $G$ on $\mu_n$, and $HK$ an abelian extension of exponent dividing $n$. Then $H=K(\root n\of D)$ for some subgroup $D\subset K^\times/K^{\times n}$, by Kummer theory. It can be checked that $HF$ is galoisian if and only if $D$ is $G$stable. When such is the case, the conjugation action of $G$ on $\mathrm{Gal}(HK)$ coming from the short exact sequence $$ 1\to\mathrm{Gal}(HK)\to\mathrm{Gal}(HF)\to G\to1 $$ is trivial if and only if $G$ acts on $D$ via $\omega$. In this situation ($H=K(\root n\of D)$ for some subgroup $D\subset(K^\times/K^{\times n})(\omega)$), a sufficient condition for $H$ to be abelian over $F$, is that the order of $G$ be prime to $n$, because then $\mathrm{Gal}(HF)=\mathrm{Gal}(HK)\times\mathrm{Gal}(KF)$. I'm sure this situation can be realised when $F=\mathbf{Q}$, for example when the finite abelian extension $K$ has odd degree $[K:\mathbf{Q}]$, $n=2$, the class group of $K$ has order ($1$ or) $2$, and $H$ is the Hilbert class field of $K$. In this case the extension $H\mathbf{Q}$ will be necessarily abelian. 


I think $K$ has to be an abelian extension of $\mathbb{Q}$ of class number one. Because if the Hilbert class field is abelian over $\mathbb{Q}$, then it is contained in a cyclotomic field, so $K$ itself is inside that cyclotomic field so $K$ is abelian, but an extension of fields inside a cyclotomic field is always ramified (not 100% sure here but I think it's OK) so the Hilbert class field cannot be a proper extension of $K$. Edit: Definitely not OK. See comments below. 

