there is no such sub-field. It is a theorem of Emil Artin that the only automorphism of the field of complex numbers of finite order is conjugate in $Aut ({\mathbb C})$ to the complex conjugation. If such a sub-field $F$ existed, then $Aut ({\mathbb C}/F)$ would have elements of order bigger than two (this is easy to prove), and hence would contradict Artin's theorem.

there is no such sub-field. [Edited] It is a theorem of Emil Artin that the only automorphism of the field of complex numbers of finite order is of order two. If such a sub-field $F$ existed, then $Aut ({\mathbb C}/F)$ would have only elements of order two and hence abelian. In particular, ${\mathbb R}/F$ would be abelian (and Galois). But $\mathbb R$ has no field automorphisms and hence $F=\mathbb R$.

I thank Peter clark for pointing out that I was attributing a wrong result to Emil Artin(!).