 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 $d=2^{\alpha }{3}^{\beta }{5}^{\gamma }$ 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 isThe 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.