Let $V\models\sf ZFC$, and let $V[r]$ be a generic extension obtained by adding one Cohen real, or equivalently $\omega$ Cohen reals.

It is clear that $\Bbb R^{V[r]}$ and $\Bbb R^V$ have the same cardinality, and that $\Bbb R^{V[r]}\setminus\Bbb R^V$ also have that cardinality.

Furthermore any generic real must be transcendental (I think, because we don't change the rationals), so $\Bbb R^{V[r]}$ is a transcendental extension of $\Bbb R^V$. But what's the transcendence degree of this extension?

It clearly cannot be finite, because adding one Cohen real adds $\omega$ pairwise generic Cohen reals, and the above argument gives us that they cannot be algebraically dependent. But is it $\aleph_0$, or is it $2^{\aleph_0}$? Maybe something else?