= Solution
An <ideal> of $R/I$ has the form $J/I$ for an <ideal> $J\supseteq I$ of $R$. If $R$ is Noetherian, write
$$
J=(a_1,\ldots,a_k).
$$
Then
$$
J/I=(a_1+I,\ldots,a_k+I),
$$
so every <ideal> of $R/I$ is finitely generated. Hence
$$
\boxed{R/I\text{ is Noetherian}}.
$$
The <Hilbert basis theorem> states that $R[X]$ is Noetherian whenever $R$ is Noetherian. The integers form a principal <ideal> domain, so every <ideal> of $\mathbb Z$ has one generator and $\mathbb Z$ is Noetherian. It follows that $\mathbb Z[X]$ is Noetherian.
For a nonsquare integer $d$, evaluation at $\sqrt d$ gives
$$
\mathbb Z[X]\longrightarrow\mathbb Z[\sqrt d],
\qquad X\longmapsto\sqrt d.
$$
It is surjective and its kernel is $(X^2-d)$. Therefore
$$
\boxed{\mathbb Z[\sqrt d]\cong\mathbb Z[X]/(X^2-d)}
$$
is Noetherian by the quotient result. This is the <noetherianity of a quadratic integer order>.
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