Nonstandard Reals in the Complex Plane [Edited so as to reflect answers and comments given so far]
Let $F$ be a real closed field. Then by 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) [ANSWERED] Is it possible for an uncountable such $F$ to be at least Lebesgue measurable ?*  
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 have a subfield isomorphic to $\Bbb{R}$.
 A: A partial answer to (b):  Consistently, yes.   Löwenheim-Skolem implies that there are non-Archimedean real closed fields of cardinality $\aleph_1$, and it is consistent that 
all sets of size $\aleph_1$ are of measure zero.   (For example, this follows from MA plus non-CH.) 
Alternatively: Take any model $V$ of set theory, let $K\in V$ be an uncountable 
non-Archimedean real closed field, and add a Cohen real $c$ to $V$.  In $V[c]$, the set of old reals has now measure zero, and $K$ is still a non-Archimedean real closed field. 
I have a feeling that an absoluteness argument should now help to get the existence of $K$ in $V$, but I cannot get it to work. I do not even know if I can get a definable (say: analytic) $K$ in $V[c]$ (though it seems to me that I can get a $K$ containing a perfect set). 
A: Is there a Cantor set $K$ in $\mathbb R$ of algebraically independent elements?  
Thinning it out, if necessary, we can assume $K^n$ has Hausdorff dimension $0$ for all $n$, and there are still other things algebraically independent of it.  Then $F_1 = \mathbb Q(K)$ is at least an analytic subfield, still of dimension zero, and its algebraic completion $F_2$ is again analytic, now isomorphic to $\mathbb C$, but not all of $\mathbb C$ and not equal to $\mathbb R$.  So $F_2$ is still of Hausdorff dimension $0$.  Of course it has a subfield $F_3$ isomorphic to $\mathbb R$.  (Probably many different ones?)  So if we now add something algebraically independent of that, specifying that it is infinitely large, say, compared to $F_3$, then can't we now do a real-completion to get something still analytic and nonarchimedean real closed?
