= Solution
A <modulus of a number field> is $\mathfrak m=\mathfrak m_0\mathfrak m_\infty$, with $\mathfrak m_0$ a nonzero integral <ideal> and $\mathfrak m_\infty$ a subset of the real places. Let $I_{\mathfrak m}$ be the <group> of <fractional ideals> prime to $\mathfrak m_0$, and let $P_{\mathfrak m}$ consist of principal <ideals> $(a)$ with $a\equiv1$ at each prime power in $\mathfrak m_0$ and $a>0$ at each selected real place. The <ray class group> is
$$
\operatorname{Cl}_{\mathfrak m}(K)=I_{\mathfrak m}/P_{\mathfrak m}.
$$
The <ray class field> $K_{\mathfrak m}$ is the finite <abelian extension> for which the <Artin map> has <kernel> $P_{\mathfrak m}$ and induces an isomorphism $\operatorname{Cl}_{\mathfrak m}(K)\cong\operatorname{Gal}(K_{\mathfrak m}/K)$. Equivalently it is the maximal <abelian extension> whose conductor divides $\mathfrak m$.
To compute its order use the <ray class exact sequence>, including signatures:
$$
\mathcal O_K^\times\longrightarrow (\mathcal O_K/\mathfrak m_0)^\times\times\{\pm1\}^{|\mathfrak m_\infty|}\longrightarrow\operatorname{Cl}_{\mathfrak m}(K)\longrightarrow\operatorname{Cl}(K)\longrightarrow1.
$$
The middle map lifts a residue and sign tuple to an element of $K$ by <weak approximation theorem>, then takes its principal <ideal> as a ray class. Different lifts differ by a principal ray factor. Its <kernel> consists exactly of tuples represented by global units. Every ordinary <ideal class> has a representative prime to the finite modulus, again by <weak approximation theorem>; this gives surjectivity of the last map. If $U_{\mathfrak m}$ denotes the <ray units>, the kernel of the unit map is $U_{\mathfrak m}$. The <Chinese remainder theorem> and counting units in each prime-power quotient give the <ray class number formula>
$$
\boxed{|\operatorname{Cl}_{\mathfrak m}(K)|=\frac{h_K\,2^{|\mathfrak m_\infty|}\,\Phi_K(\mathfrak m_0)}{[\mathcal O_K^\times:U_{\mathfrak m}]},\qquad \Phi_K(\mathfrak m_0)=N\mathfrak m_0\prod_{\mathfrak p\mid\mathfrak m_0}\left(1-\frac1{N\mathfrak p}\right).}
$$
First take $K=\mathbb Q(\sqrt{-3})$. Its <ring of integers> is $\mathbb Z[\omega]$, where $\omega=(-1+\sqrt{-3})/2$ and $\omega^2+\omega+1=0$. The <Minkowski bound for ideal classes> is $(2/\pi)\sqrt3<2$, so every <ideal class> has a representative of norm one and $h_K=1$. Its units are the six <roots of unity>. Modulo seven, the polynomial $X^2+X+1$ has distinct roots $2,4$, and therefore
$$
\mathcal O_K/(7)\cong\mathbb F_7\times\mathbb F_7,\qquad \omega\mapsto(2,4).
$$
The <unit group> of this quotient has order $36$. The unit $1+\omega$ maps to $(3,5)$, which has order six; consequently unit reduction is injective and $U_{(7)}=\{1\}$. The <ray class number formula> gives $36/6=6$.
We also identify the quotient explicitly. Conjugation interchanges the two factors, so the residue <norm> is $(x,y)\mapsto xy$. Its <kernel> has order six and equals the image of the six units, since $3\cdot5=1\pmod7$. Hence
$$
\operatorname{Cl}_{(7)}(K)\cong\mathbb F_7^\times\cong C_6.
$$
The field $M=K(\zeta_7)=\mathbb Q(\zeta_{21})$ has degree six over $K$, since the degrees over $\mathbb Q$ are twelve and two. It is <unramified> outside seven by the cyclotomic computation and the base-change argument above. At primes over seven, $K$ has completion $\mathbb Q_7$, and $M/K$ is the tame degree-six cyclotomic extension; its conductor exponent is one. Its relative arithmetic <Artin symbol> at an <ideal> $\mathfrak a$ prime to seven acts by
$$
\zeta_7\longmapsto\zeta_7^{N\mathfrak a\bmod7}.
$$
If $\mathfrak a=(a)$ with $a\equiv1\pmod{7\mathcal O_K}$, then $N_{K/\mathbb Q}(a)\equiv1\pmod7$. Thus the <Artin map> kills the principal ray subgroup. It factors through the order-six <ray class group>; the residue <norm> quotient computed above shows that it is onto the degree-six <Galois group>, so the fields coincide. This proves the explicit <ray class field of Q of square root minus three modulo seven>:
$$
\boxed{K_{(7)}=K(\zeta_7)=\mathbb Q(\zeta_{21}),\qquad[K_{(7)}:K]=6.}
$$
Now take $K=\mathbb Q(\sqrt{-6})$, and write $s=\sqrt{-6}$. Its <ring of integers> is $\mathbb Z[s]$, its <discriminant> is $-24$, and its units are $\pm1$. The quotient by $(s)$ is
$$
\mathbb Z[s]/(s)\cong\mathbb Z/6\mathbb Z,
$$
whose two units are both images of $\pm1$. The <ray class exact sequence> therefore gives $\operatorname{Cl}_{(s)}(K)\cong\operatorname{Cl}(K)$.
For its <class number>, the <Minkowski bound for ideal classes> is $(2/\pi)\sqrt{24}<4$, so each <ideal class> contains an integral <ideal> of norm at most three. The primes of norms two and three are
$$
\mathfrak p_2=(2,s),\qquad\mathfrak p_3=(3,s),\qquad\mathfrak p_2^2=(2),\quad\mathfrak p_3^2=(3),\quad\mathfrak p_2\mathfrak p_3=(s).
$$
These are the only nontrivial possibilities within the bound, and their <ideal classes> coincide and have order at most two. The <ideal> $\mathfrak p_2$ is not principal: its being principal would give an integer solution to $x^2+6y^2=2$, which is impossible. Thus $h_K=2$, the <ideal class group> is $C_2$, and the ray field is the <Hilbert class field>.
We verify its actual identity rather than stopping at the degree. Put
$$
H=K(\sqrt2)=\mathbb Q(\sqrt2,\sqrt{-3}),\qquad b=\frac{1+\sqrt{-3}}2.
$$
The two quadratic fields $\mathbb Q(\sqrt2)$ and $\mathbb Q(\sqrt{-3})$ are distinct, so $[H:\mathbb Q]=4$ and $[H:K]=2$. The integral elements $1,\sqrt2,b,\sqrt2b$ are a rational <basis>. Their trace matrix is the <Kronecker product> of the two quadratic trace matrices, with <discriminants> $8$ and $-3$. Its <determinant> is therefore $8^2(-3)^2=576$. The <discriminant-index formula for an integral lattice> shows that the field <discriminant> divides $576$. On the other hand the relative <discriminant ideal> formula gives
$$
|D_H|=|D_K|^{[H:K]}N_{K/\mathbb Q}(\mathfrak d_{H/K})=576\,N_{K/\mathbb Q}(\mathfrak d_{H/K}).
$$
The norm on the right is a positive integer, so equality is forced. Thus the relative <discriminant ideal> is the unit ideal, and $H/K$ is <unramified> at every finite prime. There are no real places of $K$. The extension is quadratic and therefore <Abelian>; its degree equals $h_K$, identifying it with the <Hilbert class field of Q of square root minus six>. Hence the <ray class field of Q of square root minus six modulo square root minus six> is
$$
\boxed{K_{(\sqrt{-6})}=K(\sqrt2)=\mathbb Q(\sqrt2,\sqrt{-3}),\qquad[K_{(\sqrt{-6})}:K]=2.}
$$
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