Weird subfields of $\Bbb{R}$

I found this problem, and I can't get an answer to it:

Prove that there are subfields of $\Bbb{R}$ that are

a) non-measurable.

b) of measure zero and continuum cardinality.

I can't seem to imagine how to construct such subfields of $\Bbb{R}$.

I expect direct constructions, but while we are waiting, let me at least offer a proof that both assertions are consistent with ZFC, since they can be obtained by forcing.

Statement (b) is true of the ground-model reals after adding a Cohen real. In the answer to this MO question, Martin Goldstern explains that the set of reals in the ground model $\mathbb{R}^V$ has measure $0$ in the forcing extension $V[c]$, but it is easy to see that it has size continuum there. So this is a subfield of $\mathbb{R}$ in the universe $V[c]$ which has measure $0$ and size continuum.

Statement (a) is true in the models mentioned in Andreas Blass' comment to my answer to this MO question, where it is explained that the set of constructible reals, which is certainly a subfield, can be non-measurable afte forcing over the constructible universe.

Perhaps one can turn both of these arguments into actual ZFC constructions by considering partially generic filters. Or perhaps there are easier direct constructions.

Rather bizarrely, here's an answer to one of the questions, but I'm not sure which. But maybe it's a useful starting point.

Pick a transcendence basis of $B$ of $\mathbb{R}$ over $\mathbb{Q}$, any $b \in B$, and look at the field $F$ generated by $B\setminus \{b\}$. Since the translates of $F$ by elements of $\mathbb{Q}b$ are pairwise disjoint, $F$ is either null or nonmeasurable.

I suspect (hope?) that this is nonmeasurable, maybe after taking algebraic closure. If this is so, then a strategy for getting a null field could be tossing out more elements of $B$.

You can try the answers (including mine) for a question at MathOverflow... https://mathoverflow.net/questions/27352/a-question-about-fields-of-real-numbers/27358#27358