Let n be an integer that is not a perfect square. We need to prove that factor ring Z[sqrt(n)]/(sqrt(d)) is isomorphic to Z_n.
Nothing in the problem is said about d. I am assuming that we have to consider 2 cases: when d is a perfect square and when it is not.
In the first case factor ring would look as an integer (?) (as Z[sqrt(n)]={m-integer}) over integer. Thus, factor ring is {m+k(k is a sort of d)|m is in Z[sqrt(n)]} which is an integer itself, so is isomorphic to Z_n.
In the second then we will have factor ring Z/Q.
I am not sure that my reasoning here is right and I understood the problem correctly.
I'm a bit confused by the statement of the problem, as it's not clear what the relationship between $n$ and $d$ is supposed to be. The best I can divine is the following:
There is a well-known ring $\mathbb Z [\sqrt n] = \mathbb Z \oplus \mathbb Z\sqrt n$. Of course, it makes sense to quotient rings by ideals, and some ideals are principal. So let's assume $\sqrt d \in \mathbb Z[\sqrt n]$. Then we can study $\mathbb Z[\sqrt n]/ (\sqrt d)$.
The first thing to note is that if $\sqrt d \in \mathbb Z[\sqrt n]$, then $\sqrt d = a + b\sqrt n$ for some $a,b\in\mathbb Z$. I assume that $d\in \mathbb Z$; then $d = a^2 + b^2 n + 2ab\sqrt n$. Assuming that $n$ is not a square, it follows that $ab = 0$, and so either $d$ is a square or $\sqrt d = b\sqrt n$.
In the first case, suppose that $\sqrt d = a$ for $a \in \mathbb Z$. Then we want to study $\mathbb Z[\sqrt n]/(a)$. Suppose that $p + q\sqrt n$ is a multiple of $a$ in $\mathbb Z[\sqrt n]$; i.e. $p + q \sqrt n = a \times (p' + q' \sqrt n)$. Then both $p,q$ are divisible by $a$. It follows that $\mathbb Z[\sqrt n]/(a) = \frac{\mathbb Z}{(a)}[\sqrt n] = \frac{\mathbb Z}{(a)} \oplus \frac{\mathbb Z}{(a)}\sqrt n$. This ring not cyclic unless $a = 1$, when it is the trivial ring.
In the second case, suppose that $\sqrt{d} = b \sqrt n$. Then $(\sqrt d)$ consists of the numbers of the form $(p + q\sqrt n)b\sqrt n = bqn + bp\sqrt n$. As an abelian group, we thus have $\frac{\mathbb Z}{(bn)} \oplus \frac{\mathbb Z}{(b)}\sqrt n$, with the obvious ring structure. This ring is cyclic only when $b = 1$, in which case it is the ring $\mathbb Z / (n)$.
From this I conclude that the intended relation between $n$ and $d$ was for them to be the same. If this question was coming from a homework sheet, my suspicion is that your instructor modified the problem after writing it, upon deciding that $d$ was a better letter to use than $n$ (or equally likely deciding in the opposite direction), but didn't make all requisite changes.