Let $p,q$ be distinct primes. Then
(1) $\dfrac{\mathbb{Z}}{p^2q}$ has exactly 3 distinct ideals.
(2) $\dfrac{\mathbb{Z}}{p^2q}$ has exactly 3 distinct prime ideals.
(3) $\dfrac{\mathbb{Z}}{p^2q}$ has exactly 2 distinct prime ideals.
(4) $\dfrac{\mathbb{Z}}{p^2q}$ has unique maximal ideal.
Generally, the ideals of $\mathbb{Z}_n$ are of the form $<d>$, where $d|n$. But how to describe all prime maximal ideals?
On
Let $I$ be an ideal of the form $<d>$, then what does the quotient $\Bbb{Z}_{p^2q}/I$ look like? If this quotient is a field then what can be said about $I$? If the number of elements of the quotient is prime then what can be said about it?
On
$\mathbb{Z}/p^2q \mathbb{Z}$ is cyclic and so abelian as a group. Its ideals correspond to subgroups, and so we can use the group theory's tools.
In an abelian group $G$ a subgroup $H$ is maximal if and only if $\frac{G}{H}$ is a simple group, i.e. if and only if $|\frac{G}{H}|$ is prime. In this case we can take $|H| = p^2$ or $|H| = pq$ and so $(4)$ is wrong.
$(1)$ is wrong because there is exactly one subgroup with cardinality $p, \ q , \ p^2 , \ pq$ because $G$ is cyclic.
Among these subgroups, only those with cardinality $pq $ and $p^2$ are primes because the quotient is a domain, and so the answer is $(3)$ .
First off, these are finite rings, and a prime ideal is maximal in a finite ring. (Proof: if $R/P$ is a domain, it's a finite domain, hence a field by Wedderburn's little theorem. Thus $P$ is maximal.)
So it suffices to find the maximal ideals.
The maximal ideals of $\Bbb Z/p^2q\Bbb Z$ are those maximal ideals of $\Bbb Z$ containing $p^2q\Bbb Z$. You probably already know the maximal ideals of $\Bbb Z$ look like $(b)$ for each prime $b\in \Bbb Z$.
Now a maximal ideal is prime, and a prime containing $p^2q$ must contain $q$ or $p$. But $(q)$ and $(p)$ are maximal ideals. So there are exactly these two prime (and also maximal) ideals.