 Solution

Put $s=\sqrt{30}$. Since $30\equiv 2(\mathrm{mod}4)$, the ring of integers is $\mathbb{Z}[s]$ and its field discriminant is $120$. All three primes ramify:For example ${\mathfrak{p}}_p^2\subseteq (p)$ and both ideals have norm $p^2$, proving equality. The Minkowski bound for ideal classes is $\sqrt{120}/2=\sqrt{30}<6$. Thus the ideal class group is generated by the classes of these prime ideals; an ideal of norm four contributes only the already principal ${\mathfrak{p}}_2^2$.

The elements $6+s$ and $5+s$ have norms $6$ and $-5$. Their ideal factorizations give ${\mathfrak{p}}_2{\mathfrak{p}}_3=(6+s)$ and ${\mathfrak{p}}_5=(5+s)$. This also explains the printed hint: $N((6+s)/2)=3/2$, a fraction, not the integer $32$. Therefore $[{\mathfrak{p}}_3]=[{\mathfrak{p}}_2]$ and $[{\mathfrak{p}}_5]=1$, leaving at most two ideal classes. The class of ${\mathfrak{p}}_2$ is nontrivial: a principal generator would have norm $\pm 2$, but $x^2-30y^2=\pm 2$ would require $x^2\equiv 2$ or $3(\mathrm{mod}5)$, impossible. Hence