[Edited so as to reflect answers and comments given so far]
Let $F$ be a real closed field. Then by Artin-Schrier Artin-Schreier theory, $F[i]$ is algebraically closed. If $F$ is further assumed to have cardinality at most continnum, then by a classical theorem of Steinitz [stating the isomorphism of any two uncountable algebraically closed fields of the same cardinality and characteristic], we can conclude that $F[i]$ is isomorphic to a subfield of $\Bbb{C}$ of complex numbers.
In particular, if $F$ is a non-archimedean real closed field of cardinality continuum, then $F[i]$ is isomorphic to $\Bbb{C}$ [The proof uses the axiom of choice in a serious way, by the way].
We can therefore conclude:
Theorem. Every nonarchimedean real closed field of power at most continuum is isomorphic to a subfield of $\Bbb{C}$.
As a special case, we may conclude that there is a subfield $F$ of $\Bbb{C}$ such that $F$ is a non-archimdean real closed field that, furthermore, has a subfield isomorphic to the field $\Bbb{R}$ of real numbers.
The above considerations allow me to state my questions.
Questions.
(a) [UNANSWERED] Is there an uncountable Borel non-Archimedean real closed field $F$ of $\Bbb{C}$?
NOTE: In his comment below Dave Marker asks whether this question has a negative answer if we further assume that $F[i]=\Bbb{C}$, then Gerald Edgar pointed out in his comment that this is indeed the case; based on a result that appears in a joint paper of his with Chris Miller.
(b) If the answer to (a) is negative, is [ANSWERED] Is it possible for an uncountable such $F$ to be at least Lebesgue measurable ? [if so, the measure has to be 0 by Steinhaus' theorem]. *
NOTE. (b) has been answered. First Martin Goldstern pointed out that (b) follows from $MA + \lnot CH$; and that (b) is also true in any universe of set theory obtained by adding a Cohen real. Then Gerald Edgar pointed out that (b) is provable outright in $ZFC$ [see the answers below].
(c) [UNANSWERED] If the answer to (a) is positive, does the answer change if we insist for $F$ to satisfy have a subfield isomorphic to $\Bbb{R}$.
