Consider the ring $R_{\omega}:=\mathbb{Z} \oplus \omega\mathbb{R}[\omega]$ which is contained in $\mathbf{Oz}$. We have $\mathbb{R} \not\subseteq R_{\omega}$ and $\operatorname{Frac}(R_{\omega})\supseteq \mathbb{R}$.

Consider a ring $R_0$ given by the answer linked by Kevin Casto. The ring $R_0$ is archimedean, so any smallest ordered ring $R$ you are looking for must be archimedean. Since $R$ cannot be just $\mathbb{Z}$, it contains some element lying strictly between $0$ and $1$. But then $R$ cannot embed in $R_{\omega}$ for which this isn't the case. So such an ordered ring $R$ does not exist.

Assume for contradiction that such a ring $R$ exists.

Consider the ring $R_{\omega}:=\mathbb{Z} \oplus \omega\mathbb{R}[\omega]$ which is contained in $\mathbf{Oz}$. We have $\mathbb{R} \not\subseteq R_{\omega}$ and $\operatorname{Frac}(R_{\omega})\supseteq \mathbb{R}$. Since $R_{\omega}$ is discretely ordered, so must be $R$.

Consider a ring $R_0$ given by the answer linked by Kevin Casto. Since $R_0$ is archimedean, so must be $R$. 

But $\mathbb{Z}$ is the only discretely ordered archimedean ordered ring. So we must have $R=\mathbb{Z}$, which cannot be.