Often I have encountered questions like :
How many rings of order 4 are there upto isomorphism?
Often the solution involved brute-force treatments as checking the multiplication tables.
But it is possible to check all possible multiplication tables if the order of the ring is small.But it becomes very difficult when it comes to any arbitrary number $n$.
So,does there exist any general formulation to find the number of rings of order $n$ upto isomorphism of rings?
According to this paper
Singmaster, David; Bloom, D. M. (October 1964), "E1648", American Mathematical Monthly, 71 (8): 918–920
there are 11 rings of order 4. According to this question, there are 4 unitary rings of order 4.
The number of rings with $n$ elements is sequence A027623 in the The On-Line Encyclopedia of Integer Sequences. There is no known general formula
I will try to answer your question for the case of order $p^2$, where $p$ is a prime, and for unitary commutative rings $R$.
Note first that we have a unique homomorphism $f:\mathbb{Z}\to R$, sending $1$ to $1$.
If the image of $f$ is $R$, then $R\cong \mathbb{Z}/\ker(f)$, hence $R\cong \mathbb{Z}/p^2\mathbb{Z}$ as rings.
If the image is not $R$, it must have order $p$, hence $$\mathbb{F}_p:=\mathbb{Z}/p\mathbb{Z} \hookrightarrow R$$ which makes $R$ an $\mathbb{F}_p$-vector space, necessarily of dimension $2$.
If $R$ is a field, then, by a general theorem on existence and unicity of finite fields of order $p^n$, then $R\cong \mathbb{F}_{p^2}$, the unique field of order $p^2$.
If $R$ is an integral domain, since any finite domain is a field , we are reduced to the previous case.
If $R$ is not an integral domain, either there exists $a\ne b\in R$, $a\ne 0 \ne b$, such that $a\cdot b=0$, or there exists $a\in R$, $a\ne 0$, such that $a^2=0$.
Consider first the second case: then, consider the morfism of rings $g:\mathbb{F_p}[x]\to R$ sending $p(x)$ to $p(a)$. Clearly $g(\mathbb{F_p}[x])= R$, since $\langle 1,a\rangle _{\mathbb{F}_p} =R$ already as vector spaces (since $a\notin \mathbb{F}_p$), and $\ker(x)\supset (x^2)$. But $\mathbb{F_p}[x]/(x^2)$ has $p^2$ elements, hence $R\cong \mathbb{F_p}[x]/(x^2)$.
We are left with the case that there exists $a\ne b\in R$, $a\ne 0 \ne b$, such that $a\cdot b=0$, but $c^2\ne 0$ for all $c\in R$, $c\ne 0$. We will see that $R\cong \mathbb{F_p}\times \mathbb{F}_p$. The idea is that $\langle a, b\rangle _{\mathbb{F}_p} =R$, since they are both linearly independent over $\mathbb{F}_p$: if $a=n\cdot b$, $n\in \mathbb{F}_p$, $n\ne 0$, then $0=a\cdot b=n\cdot a^2$, hence $a^2=0$, which we suppose is not true. Now, there exists $n,m\in \mathbb{F}_p$, $n\ne 0 \ne m$, such that $1=na+mb$. After changing $a$ and $b$ for $na$ and $mb$, we can suppose $a+b=1$. But then $a^2=a(1-b)=a-ab=a$, and similarly for $b$. One cheks that the bijection $\psi: \mathbb{F_p}\times \mathbb{F}_p\to R$ sending $\psi(i,j)=i\cdot a+j\cdot b$ is a morphism of rings, so an isomorphism.
Hence, there are $4$ unitary commutative rings with $p^2$ elements: $$ \mathbb{Z}/p^2\mathbb{Z},\ \mathbb{F}_{p^2},\ \mathbb{F_p}\times \mathbb{F}_p \text{ and } \mathbb{F_p}[x]/(x^2).$$
p.s. After writting this answer I learn that all unitary rings of order $p^2$ are commutative. So previous argument covers all unitary rings.
