Let $R$ be a domain with fraction field $Q$, and let $A$, $C$ be $R$-modules. Then, if $A$ is a vector space over $Q$, $Ext_{R}^{n}(C,A)$ is a vector space over $Q$.
Let $0\rightarrow A\xrightarrow{\eta}E^0\xrightarrow{d^0}E^1\xrightarrow{d^1}E^2\rightarrow\cdots$ be an injective resolution. Then, by definition, $Ext_{R}^{n}(C,A)=\mathrm{ker}\ d^{n}_{*}/\mathrm{im}\ d^{n-1}_{*}$, where $d^{n}_{*} : Hom_{R}(C,E^n)\rightarrow Hom_{R}(C,E^{n+1})$.
I guess $Hom_{R}(C,E^n)$ should be a vector space over $Q$ but I am not sure.
The statement is a bit sloppy: $A$ should not be just a $Q$-vector space, but rather it should be a $Q$-vector space so inducing the structure of $R$-module.
You should rather use a projective resolution of $C$, because if $M$ is any $R$-module, you can give $\operatorname{Hom}_R(M,A)$ a structure of $Q$-vector space which is functorial: if $M\to N$ is a morphism of $R$-modules, then the induced map $\operatorname{Hom}_R(N,A)\to\operatorname{Hom}_R(M,A)$ is a morphism of $Q$-vector spaces.