You can use basic exact sequences of cohomology to compute that number. In what follows, to keep notations light, I will omit the cyclic groups, so for instance $H^2=H^2(C_p,\mathbb{Z}/p\mathbb{Z})$. What you want to compute is the order of $Z^2$.
We have $0\to B^2\to Z^2\to H^2\to 0$ and $0\to Z^1\to C^1\to B^2\to 0$, so we are done if we can compute the order of $H^2$, $Z^1$ and $C^1$. Explicitly, $|Z^2|=\frac{|H^2|\cdot |C^1|}{|Z^1|}$.
Now $C^1$ is just the set of functions $C_p\to \mathbb{Z}/p\mathbb{Z}$, so $|C^1|=p^p$. Also $Z^1$ is the set of group morphisms $C_p\to \mathbb{Z}/p\mathbb{Z}$, so $|Z^1|=p$.
Finally, the most complicated is $H^2$, but the cohomology of cyclic groups is very well-known, and $H^2=\mathbb{Z}/p\mathbb{Z}$, so $|H^2|=p$.
In the end, $|Z^2|=p^p$.
EDIT: What is written above is the "correct" way to compute the number you are looking for. But since you want a computation that does not explicitly involve cohomology, here is a completely elementary proof.
We can write an explicit bijection between $Z^2$ and the set $X$ of functions $C_p\to \mathbb{Z}/p\mathbb{Z}$: to each $c\in Z^2$ we associate
$$\varphi_c: x\mapsto c(x,1),$$
and to each $\varphi\in X$ we associate
$$c_\varphi: (x,y)\mapsto \varphi(0) + \sum_{i=0}^{y-1}(\varphi(x+i)-\varphi(i))$$
where we assume that $1\leqslant y\leqslant p$.
The trickiest part is to show that $c_\varphi$ is actually a $2$-cocycle. We will do that at the end. For now let's assume this is true, and show that the above associations are mutually inverse bijections. It is immediate that $c_\varphi(x,1)=\varphi(x)$, so in other words $\varphi_{c_\varphi}=\varphi$. It is a little less obvious that
$$ \varphi_c(0) + \sum_{i=0}^{y-1}(\varphi_c(x+i)-\varphi_c(i)) = c(x,y)$$
(in other words $c_{\varphi_c}=c$), but not very hard to check either. We can proceed inductively: if $y=1$, then the equality just means that $\varphi_c(0)+\varphi_c(x)-\varphi_c(0)=c(x,1)$, which is the definition of $\varphi_c$. Now if the equality is true for some $1\leqslant y<p$, then using the cocycle condition:
$$\begin{align}
c(x,y+1) & = c(x,y)+c(x+y,1)-c(y,1) \\
& = c(x,y)+\varphi_c(x+y)-\varphi_c(y) \\
& = \varphi_c(0) + \sum_{i=0}^{y-1}(\varphi_c(x+i)-\varphi_c(i))+\varphi_c(x+y)-\varphi_c(y) \\
&= \varphi_c(0) + \sum_{i=0}^{y}(\varphi_c(x+i)-\varphi_c(i))
\end{align}$$
as we wanted. In the end $c\mapsto \varphi_c$ and $\varphi\mapsto c_\varphi$ are inverse of each other, which shows that $|Z^2|=|X|=p^p$.
It remains to show that $c:=c_\varphi$ is a $2$-cocycle. We want to establish that if $1\leqslant x,y,z\leqslant p$
$$c(y,z)+c(x,y+z) = c(x+y,z)+c(x,y).$$
First, suppose $y+z\leqslant p$. Then on the left-hand side, we have
$$2\varphi(0)+\sum_{i=0}^{z-1}\varphi(y+i)-\varphi(i) + \sum_{i=0}^{y+z-1}\varphi(x+i)-\varphi(i)$$
and on the right-hand side:
$$2\varphi(0)+\sum_{i=0}^{z-1}\varphi(x+y+i)-\varphi(i) + \sum_{i=0}^{y-1}\varphi(x+i)-\varphi(i).$$
So clearing the terms that appear on both sides, the equality amounts to
$$\sum_{i=0}^{z-1}\varphi(y+i) + \sum_{i=y}^{y+z-1}\varphi(x+i)-\varphi(i) = \sum_{i=0}^{z-1}\varphi(x+y+i),$$
which is easy to check since $\sum_{i=y}^{y+z-1}\varphi(x+i)-\varphi(i) = \sum_{i=0}^{z-1}\varphi(x+y+i)-\varphi(y+i)$.
Finally, suppose $y+z>p$; let us write $y+z=p+r$, with $r\geqslant 1$. Note that $r\leqslant y$. Then the left-hand side is
$$2\varphi(0)+\sum_{i=0}^{z-1}\varphi(y+i)-\varphi(i) + \sum_{i=0}^{r-1}\varphi(x+i)-\varphi(i)$$
and the right-hand side is the same:
$$2\varphi(0)+\sum_{i=0}^{z-1}\varphi(x+y+i)-\varphi(i) + \sum_{i=0}^{y-1}\varphi(x+i)-\varphi(i).$$
If we simply the terms appearing on both sides, we get the equality
$$\sum_{i=0}^{z-1}\varphi(y+i) = \sum_{i=0}^{z-1}\varphi(x+y+i) + \sum_{i=r}^{y-1}\varphi(x+i)-\varphi(i).$$
The left-hand side can be rewritten as
$$\sum_{i=y}^{p-1}\varphi(i)+\sum_{i=0}^{r-1}\varphi(i)$$
and the right-hand side as
$$\sum_{i=y}^{p-1}\varphi(x+i) + \sum_{i=0}^{r-1}\varphi(x+i) + \sum_{i=r}^{y-1}\varphi(x+i) - \sum_{i=r}^{y-1}\varphi(i) = \sum_{i=0}^{p-1}\varphi(i)-\sum_{i=r}^{y-1}\varphi(i)$$
and now both sides are easily seen to be equal.
