= Solution
Here $n=2$, $r_2=1$, and $d_K=-47$, so the <Minkowski bound for ideal classes> is
$$
\frac2\pi\sqrt{47}<5.
$$
Every ideal class therefore has a representative of norm at most four. Such classes are generated by the prime ideals above $2$ and $3$. Put
$$
x=[\mathfrak p_2],
\qquad
y=[\mathfrak p_3]
$$
in additive notation. Since conjugate prime ideals represent inverse classes, the two principal-ideal factorizations from part (b) give
$$
2x+y=0,
\qquad
x-2y=0.
$$
Thus $y=-2x$ and $5x=0$, so the class group is generated by $x$ and has order dividing five.
It is not trivial. If $\mathfrak p_2$ were principal, there would be an element of norm two, but
$$
N(a+b\omega)=a^2+ab+12b^2
=\left(a+\frac b2\right)^2+\frac{47}{4}b^2
$$
cannot equal two for integers $a,b$. Therefore $x\ne0$. Since five is prime,
$$
\boxed{\operatorname{Cl}(\mathcal O_K)\cong C_5.}
$$
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