 Solution

Write $r=\sqrt{10}$. The ideals are ${\mathfrak{p}}_2=(2,r)$, ${\mathfrak{p}}_3=(3,r+1)$ and ${\overline{\mathfrak{p}}}_3=(3,r-1)$, with residue fields of orders 2, 3 and 3 respectively. Thus their ideal norms are 2, 3 and 3.

The generators of ${\mathfrak{p}}_2^2$ are $4,2r,10$. They are multiples of 2, and the integer combination of 4 and 10 giving 2 proves ${\mathfrak{p}}_2^2=(2)$. Similarly ${\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$ is generated by $9,3r+3,3r-3,r^2-1=9$. The difference of the middle generators is 6; together with 9 this generates 3. Therefore the product is $(3)$.

The ideals ${\mathfrak{p}}_2$ and ${\mathfrak{p}}_3$ are coprime, so their product equals their intersection. The element $4+r$ belongs to both, hence $(4+r)\subseteq {\mathfrak{p}}_2{\mathfrak{p}}_3$. Both ideals have norm 6, so equality holds. Likewise $2+r$ belongs to ${\mathfrak{p}}_2$ and ${\overline{\mathfrak{p}}}_3$, and its norm is $-6$. ThusThe discriminant is 40. The Minkowski bound for ideal classes in a real quadratic field is $\sqrt{40}/2=\sqrt{10}<4$, so every ideal class has an integral representative of norm 1, 2 or 3. The only possible prime ideals with those norms are the ones just listed. Their classes satisfy $[{\mathfrak{p}}_2{]}^2=1$ and $[{\mathfrak{p}}_3]=[{\overline{\mathfrak{p}}}_3]=[{\mathfrak{p}}_2]$. Thus there are at most two classes. The ideal ${\mathfrak{p}}_2$ is not principal, since a generator would have norm $\pm 2$, already ruled out. Therefore the class number is exactly two.