For a field $K$ to be a field of fractions of an integral domain $R$, it is necessary and sufficient that (i) $R \subset K$ and (ii) For all $k \in K$, there exists $s \in R$ with $s \neq 0$ such that $k \cdot s \in R$ (i.e. if we write $r = k \cdot s$ then $k = \frac{r}{s}$).
Let us verify the claim. As you mention in the comments, the definition of a field of fractions is being the smallest field $K$ such that $R \subset K$. So let's assume that $R \subset K$ and show: $$K \text{ is the field of fractions of } R \Leftrightarrow \text{Condition (ii) given above}.$$
($\Rightarrow$) Suppose that $K$ is the field of fractions of $R$. Define $$K'= \left\{k \in K : \text{ there exists } r \in R \text{ such that } r \neq 0 \text{ and } k \cdot r \in R\right\}.$$
Then one can check (you should do so!) that $K'$ is a field. Certainly $K'$ contains $R$. Therefore since $K$ is the field of fractions (the smallest field that contains $R$) and $K' \subset K$, we get that $K' = K$. Therefore $K$ satisfies condition (ii).
($\Leftarrow$) Now let $K$ be a field with $R \subset K$ and $K$ satisfying condition (ii). We will show $K$ is the field of fractions of $R$. Denote the field of fractions of $R$ by $F(R)$. To prove the claim, we will show that for any $k \in K$, $k \in F(R)$. Indeed, then since $F(R)$ is the smallest field containing $R$, and $K \subset F(R)$, it must be that $K = F(R)$.
So let $k \in K$, and by condition (ii) there exists $s \in R$ with $s \neq 0$ such that $k \cdot s \in R$. So write $k \cdot s = r$, and hence (in $K$) we have $k = rs^{-1}$. But $r \in F(R)$ and $s^{-1} \in F(R)$ so therefore $k = rs^{-1} \in F(R)$. This proves that $K \subset F(R)$, as claimed.