= Solution
Dedekind's criterion factors $(p)$ according to the factorization of $X^3-7$ modulo $p$ (the monogenic assumption removes the index obstruction). With $\alpha=\sqrt[3]7$,
$$
(2)=(2,\alpha-1)(2,\alpha^2+\alpha+1),\quad
(3)=(3,\alpha-1)^3,
$$
$$
(5)=(5,\alpha-3)(5,\alpha^2+3\alpha+4),\quad
(7)=(\alpha)^3.
$$
Minkowski's bound is below 11, so classes are generated by prime <ideals> over $2,3,5,7$. Relations from the displayed factorizations and principal <ideals> reduce them to $\mathfrak p=(3,\alpha-1)$ with $\mathfrak p^3$ principal. Part (a) shows $\mathfrak p$ is not principal, so $\operatorname{Cl}(K)\simeq\mathbb Z/3\mathbb Z$.
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