$\mathbb{Q}$ is the smallest Archimedean ordered field. That is, any Archimedean ordered field (including $\mathbb{R}$) contains $\mathbb{Q}$ as a subfield.
So does there exist a maximal Archimedean ordered field? That is, does there exist an Archimedean ordered field $\mathbb{F}$ such that any Archimedean ordered field (including $\mathbb{R}$) must be contained in $\mathbb{F}$?
Yes, that field is $\mathbb{R}$. Given any Archimedean ordered field $F$, there is an embedding $\mathbb{Q}\to F$. For any $x,y\in F$ with $x<y$, the Archimedean property implies there are $a,b\in \mathbb{Q}$ such that $a<x<y<b$, and there is also $c\in\mathbb{Q}$ such that $1/(y-x)<c$. Let $n$ be the largest integer such that $a+n/c\leq x$. Then since $1/c<y-x$, we have $x<a+(n+1)/c<y$.
That is, we have shown that for any $x<y$ in $F$, there is some rational $d=a+(n+1)/c$ such that $x<d<y$. In particular this means that every $x\in F$ is uniquely determined by the set of rational numbers that are less than $x$. For each $x\in F$, let $\varphi(x)\in\mathbb{R}$ be the supremum of the set $\{a\in\mathbb{Q}:a<x\}$. By the discussion above, $\varphi:F\to\mathbb{R}$ is injective, and it is clearly order-preserving. It is also straightforward (though tedious) to prove that $\varphi$ preserves addition and multiplication (this is similar to the verification that the construction of $\mathbb{R}$ by Dedekind cuts really is a field).