= Solution
If $M=0$, take the empty filtration. Otherwise a nonzero quotient $Q=M/M_i$ is a finitely generated <module>. Because $R$ has the <ascending chain condition> on ideals, the nonempty collection of annihilators of nonzero elements of $Q$ has a maximal element. Part (b), applied to $Q$, supplies $q\ne0$ with a <prime ideal> $P=\operatorname{Ann}(q)$.
The map $R\to Rq$, $r\mapsto rq$, is surjective with kernel $P$. The <first isomorphism theorem> gives $Rq\cong R/P$. Let $M_{i+1}$ be the inverse image of $Rq$ under $M\to Q$. Then
$$
M_i\subsetneq M_{i+1},\qquad \boxed{M_{i+1}/M_i\cong R/P.}
$$
Repeat whenever the quotient remains nonzero. An infinite repetition would give a strictly ascending chain of <submodules> of the <Noetherian module> $M$, contradicting part (a). Thus the process stops after finitely many steps, necessarily at $M$.
\b[This constructs the required <prime filtration>] $0=M_0\subsetneq\cdots\subsetneq M_l=M$, with each successive quotient $R/P_i$ for a prime $P_i$.
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