It is a well-known fact that the category of abelian groups has enough injectives, hence actually any abelian group has an injective resolution. Furthermore, the global dimension of the ring $\mathbb{Z}$ is $1$, so actually these injective resolutions are bounded (and, actually, very short). For instance, we may consider the following injective resolution of the free abelian group $\mathbb{Z}$

$$ 0 \to \mathbb{Z} \to \mathbb{Q} \to \mathbb{Q}/\mathbb{Z} \to 0 \to \cdots $$

Where $\mathbb{Z} \to \mathbb{Q}$ is the canonical inclusion and $\mathbb{Q} \to \mathbb{Q}/\mathbb{Z}$ is the canonical projection map induced by the quotient. As you have already pointed out, $\mathbb{Q},\mathbb{Q}/\mathbb{Z}$ are injective abelian groups and it is trivial that the sequence is exact. I don’t know whether this injective resolution is a “good example” but I would argue that is a rather canonical example of an injective resolution in the literature. For a nice application of this resolution, you may look up at *Basic Homological Algebra, Osborne* **3.4.Example 11** in which uses this resolution to prove that $\mathrm{Ext}^1(\mathbb{Q},\mathbb{Z}) \cong \mathbb{R}$ as abelian groups. 

I am sorry I lack enough background to provide you with a more geometrical example. Nonetheless, I would like to provide also another elementary example with a rather different nature, which I believe to be very illustrative. We are going to consider now modules over the ring $\mathbb{Z}/(4)$. It is easy to see using Baer’s criterion that the free $\mathbb{Z}/(4)$-module $\mathbb{Z}/(4)$ is also injective. You can now verify that the following complex is, indeed, an injective resolution for the $\mathbb{Z}/(4)$-module $\mathbb{Z}/(2)$: 

$$ 0 \to \mathbb{Z}/(2) \xrightarrow{\cdot2} \mathbb{Z}/(4) \xrightarrow{\cdot2} \mathbb{Z}/(4) \xrightarrow{\cdot 2} \cdots $$

Something interesting is that this resolution cannot be “improved” replacing it by a finite injective resolution. A way of proving this fact is noticing that $\mathrm{Ext}^n_{\mathbb{Z}/(4)}(\mathbb{Z}/(2),\mathbb{Z}/(2))$ never vanishes. It is quite remarkable that, even though $\mathbb{Z}/(4)$-modules can be though as abelian groups which have a trivial action with respect to the ideal $(4)$, the homological behaviour of both rings is incredibly different. One can see a similar behaviour with the rings $k[T]/(T^k),k \geq 2$. 

I hope my answer has given you some insight. I would be happy to fill in the necessary gaps to help your understanding.