I realize that this question may be very wild and unusual.
Is every Hilbert class field, contained in a Hilbert class field of an imaginary quadratic field?
I think the answer is probably no. If the answer is no, can anyone give an explicit counter-example?
Give a counter-example for the above statement: Give a field whose Hilbert class field is not contained in the Hilbert class field of any imaginary quadratic field.
No. Let $K$ be an imaginary quadratic field, and let $H$ be its Hilbert class field. Then $\mathrm{Gal}(H/K)$ is abelian, since it is isomorphic to the class group of $K$.
Since $K/\mathbb Q$ is Galois, $H$ is a Galois extension of $\mathbb Q$ whose Galois group sits inside a short exact sequence $$1\to A=\mathrm{Gal}(H/K)\to \mathrm{Gal}(H/\mathbb Q)\to B = \mathrm{Gal}(K/\mathbb Q)\to 1,$$ where $A$ is an abelian group and $B\cong C_2$.
There are plenty of fields that cannot be a subfield of $H$ and, if $L$ is not a subfield of $H$, neither is its Hilbert class field.
An explicit example is the splitting field of $f(x) = x^5 + x +1$ over $\mathbb Q$, whose Galois group $S_5$ is not solvable.