I'm trying to find an abelian group $B$ such that $\operatorname{Ext}^{1}_{\mathbb{Z}}(\mathbb{Q},B)$ is non-zero. My first guess was just to choose $B=\mathbb{Z}$. Using the following argument, I deduced that $\operatorname{Ext}^{1}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z})=\{0\}$, however I then read this question, which says in fact it is non-zero, so I'm assuming there's something wrong with my argument:
First, take the injective resolution $0\rightarrow{\mathbb{Z}}\rightarrow{\mathbb{Q}}\rightarrow{\mathbb{Q}/\mathbb{Z}}\rightarrow{0}$ of $\mathbb{Z}$, form the deleted resolution $0\rightarrow{\mathbb{Q}}\rightarrow{\mathbb{Q}/\mathbb{Z}}\rightarrow{0}$ (no longer exact) and apply the $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Q},\bullet)$ functor to the deleted resolution to obtain the non-exact sequence
$0\rightarrow{\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Q})}\rightarrow{\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Q}/\mathbb{Z})\rightarrow{0}}$
So, as far as I can see, $\operatorname{Ext}^{1}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Z})$ is the quotient of the kernal of the zero map from $\operatorname{Hom}_{\mathbb{Z}}(\mathbb{Q},\mathbb{Q}/\mathbb{Z})$ by the image of the surjective map displayed in the functored sequence above, hence is zero. I'm guessing I have made a mistake either with the way I'm interpreting the functored sequence or with the definition of $\operatorname{Ext}$. Can anyone help me understand where I have gone wrong?