= Solution
The zero <ideal> is a <prime ideal> because $R$ is an <integral domain>. Each of $(2),(3),(5)$ is a <prime ideal>: for $p$ among these three, reduction $a/b\mapsto\overline a\,\overline b^{-1}$ is a surjective <ring homomorphism> $R\to\mathbb F_p$ with <kernel of a ring homomorphism> $(p)$, so its <quotient ring> is a <field>.
Any other nonzero proper <ideal> in (c) has generator $d=2^\alpha3^\beta5^\gamma$ with total exponent at least two. Choose one <prime factor> $p$ and put $e=d/p$. Then $pe=d\in(d)$ but neither $p$ nor $e$ belongs to $(d)$, as seen from the required <prime> exponents. Such an <ideal> is not <prime>. Thus with the standard proper-<ideal> convention the answer is
$$
\boxed{(0),(2),(3),(5).}
$$
\b[The definition printed in this paper omits properness.] Literally under that definition $R$ itself also satisfies the implication, so its literal complete list is the preceding four <ideals> together with $\boxed{R}$. Both interpretations are thereby accounted for.
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