As a graded $\mathbb{Z}$module, the structure of the group cohomology $H^{*}(\mathbb{Z}/n\mathbb{Z};\mathbb{Z})$ is extremely wellknown. Yet, I am having difficulty finding a reference concerning its cup product structure. I assume this is also wellknown, but I would appreciate a reference containing a precise statement of the $\mathbb{Z}$algebra structure.
See Example 3.41, p. 251 of Hatcher's Algebraic Topology. This example computes the cohomology ring of finite Lens spaces. Since the infinite Lens space is a $K(\mathbb{Z}/n,1)$, and is a increasing union of finite Lens spaces, the same computation holds for the infinite Lens spaces. Edit: My original answer was incomplete, since Example 3.41 computes the ring $H^*(\mathbb{Z}/n,\mathbb{Z}/n) \cong \mathbb{Z}/n [ \alpha, \beta]/(\alpha^2k\beta)$ (where $k=0$ if $n$ is odd, and $k=n/2$ if $n$ is even, see Hatcher's comment below, and this is an gradedcommutative ring with $\alpha$ of degree $1$, and $\beta$ of degree $2$), whereas you want $H^*(\mathbb{Z}/n,\mathbb{Z})$. We have $H^{2i}(\mathbb{Z}/n,\mathbb{Z}) \cong \mathbb{Z}/n$ for $i>0$, $H^0(\mathbb{Z}/n,\mathbb{Z})\cong \mathbb{Z}$, and $H^{2i+1}(\mathbb{Z}/n,\mathbb{Z}) \cong 0$ by universal coefficients (or Poincare duality). Let $\beta\in H^2(\mathbb{Z}/n,\mathbb{Z})$ be a generator, then the map from $H^*(\mathbb{Z}/n,\mathbb{Z})\to H^*(\mathbb{Z}/n,\mathbb{Z}/n)$ is a ring isomorphism for even degrees above zero, so we see that $H^*(\mathbb{Z}/n,\mathbb{Z})\cong \mathbb{Z}[\beta]/(n\beta)$, where $\beta$ has degree $2$ (and is torsion of order $n$). 


I don t recall seeing the computation, but you can almost immediately obtain a description as follows. Let me write $C$ for the infinite cyclic group, so as not to have too many $\mathbb Z$s. Consider the Lyndon–Hochschild–Serre spectral sequence corresponding to the extension of groups $$0\to C\to C\to\mathbb Z/n\mathbb Z\to0$$ whose $E_2$ page looks like $H^p(\mathbb Z/n\mathbb Z,H^q(C,\mathbb Z))\Rightarrow H^\bullet(C,\mathbb Z)$. It is very easy to see that $H^q(C,\mathbb Z)$ is $\mathbb Z$ for $q\in\lbrace 0,1\rbrace$ and zero otherwise. It follows now from convergence of the spectral sequence and the distribution of the zeroes in the $E_2$ page (we only have two rows) and in the limit, that the differential $d_2^{p,1}:H^p(\mathbb Z/n\mathbb Z,\mathbb Z)\to H^{p+2}(\mathbb Z/n\mathbb Z,\mathbb Z)$ is surjetive for $p=0$ and an isomorphism for $p>0$. Now $d_2$ is given by the cup product with the class $\zeta=d_2^{0,1}(1)\in H^2(\mathbb Z/n\mathbb Z,\mathbb Z)$ (Here the $1$ is that of $H^0(\mathbb Z/n\mathbb Z,\mathbb Z)$) This is enough to get the whole ring structure. Indeed, the spectral sequence degenerates at $E_2$, so gives us an exact sequence (values in $\mathbb Z$ everywhere) $$0\to H^1(\mathbb Z/n\mathbb Z)\to H^1(C)\to H^0(\mathbb Z/n\mathbb Z)\xrightarrow{\zeta\cup(\mathord)}H^2(\mathbb Z/n\mathbb Z)\to0$$ and isomorphisms $$\zeta\cup(\mathord):H^p(\mathbb Z/n\mathbb Z))\to H^{p+2}(\mathbb Z/n\mathbb Z)$$ for all $p\geq1$. We know that $H^1(\mathbb Z/n\mathbb Z)$ is torsion, and according to the exact sequence it is also a subgroup of $H^1(C)=\mathbb Z$, so $H^1(\mathbb Z/n\mathbb Z)$, and along with it all the odddegree groups, is zero. The map $H^1(C)\to H^0(\mathbb Z/n\mathbb Z)$ is easily computed (it is one of the maps appearing in the usual «$5$term sequence», as in HiltonStammbach, Th. VI.8.1) to be multiplication by $n$, so $H^2(\mathbb Z/n\mathbb Z)\cong \mathbb Z/n\mathbb Z$ generated by the class $\zeta$. The isomorphisms above then imply that $H^{2p}(\mathbb Z/n\mathbb Z)$ is cyclic of order $n$, generated by $\zeta^p$. 


Well, it's kind of a homework exercise. In particular, in Ken Brown's "Cohomology of Groups" it's Exercise V.3.2 which recommends using diagonal approximation. Here is my solution for completeness, using only the basic machinery: Let $G=\langle t\rangle$ be a finite cyclic group of order $n$ and let $F$ be the periodic resolution $\cdots\rightarrow\mathbb{Z}G\stackrel{t1}{\rightarrow}\mathbb{Z}G\stackrel{N}{\rightarrow}\mathbb{Z}G\stackrel{t1}{\rightarrow}\mathbb{Z}G\stackrel{\varepsilon}{\rightarrow}\mathbb{Z}\rightarrow 0$, where $N=\sum_{i=0}^{n1} t^i$ is the norm element. Let $\Delta:F\rightarrow F\otimes F$ be the diagonal approximation map whose $(p,q)$component $\Delta_{pq}:F_{p+q}\rightarrow F_p\otimes F_q$ is given by $\Delta_{pq}(1) =$ Consider the cohomology groups $H^{2r}(G,M)\cong M^G/NM$ and $H^{2r+1}(G,M')\cong Ker(N:M'\rightarrow M')/IM'$ where $I=\langle t  1\rangle$ is the augmentation ideal of $G$. 

