This question is inspired by this answer: There it is claimed that 'all archimedean ordered fields are subfields of reals'. Also, that there is a non-archimedean ordered field for every cardinality greater than $|\Bbb R|$.
How many archimedean ordered fields are there? I mean, are they finite?
I think I read somewhere that $\Bbb Q$ is smallest ordered field. Also, Dedekind completion of $\Bbb Q$ is $\Bbb R$. Is this claim correct that 'if $\Bbb F$ is an archimedean ordered field, then Dedekind completion of $\Bbb F$ is $\Bbb R$?'
Up to isomorphism, there are $2^{2^{\aleph_0}}$ different archimedean ordered fields. Indeed, every subfield of $\mathbb{R}$ is an archimedean ordered field, and no two different subfields of $\mathbb{R}$ are isomorphic as ordered fields (since each element is determined by the set of rational numbers less than it, which is preserved by any isomorphism of ordered fields). Now let $B\subset\mathbb{R}$ be a set of $2^{\aleph_0}$ algebraically independent real numbers. For any $S\subseteq B$, $\mathbb{Q}(S)$ is a subfield of $\mathbb{R}$, and these subfields are all distinct. This gives $2^{2^{\aleph_0}}$ non-isomorphic archimedean ordered fields. Conversely, there can't be more than $2^{2^{\aleph_0}}$ such fields since there are only $2^{2^{\aleph_0}}$ subsets of $\mathbb{R}$.
As for your second question, since $\mathbb{R}$ is Dedekind-complete and any archimedean ordered field is a dense subset of $\mathbb{R}$, the Dedekind-completion of any archimedean ordered field is $\mathbb{R}$ (up to isomorphism).