The answer is no. If $\mathbb{R}^V$ is measurable in a forcing extension $V[G]$ having new reals, then the measure must be $0$. The point is that every new real $x$ in $V[G]$ but not in $V$ is transcendental over $\mathbb{R}^V$, since one cannot add algebraic numbers by forcing. It follows that the translates of $\mathbb{R}^V$ by the powers of $x$ are disjoint. Thus, a Vitali-style argument with wrapped translations of the unit interval of $V$ shows that if $\mathbb{R}^V$ is measurable, it must have measure zero.

Gunter Fuchs and I had made this observation in connection with Sebastian's question Probabilities independent of ZFC?, which is very much related to your question.

The answer is no. If $\mathbb{R}^V$ is measurable in a forcing extension $V[G]$ having new reals, then the measure must be $0$. The point is that every new real $x$ in $V[G]$ but not in $V$ is transcendental over $\mathbb{R}^V$, since one cannot add algebraic numbers by forcing. It follows that the translates of $\mathbb{R}^V$ by the powers of $x$ are disjoint. Thus, a Vitali-style argument with wrapped translations of the unit interval of $V$ shows that if $\mathbb{R}^V$ is measurable, it must have measure zero.

Gunter Fuchs and I had made this observation in connection with Sebastian's question Probabilities independent of ZFC?, which is very much related to your question.

The answer is no. If $\mathbb{R}^V$ is measurable in a forcing extension $V[G]$ having new reals, then the measure must be $0$. The point is that every new real $x$ in $V[G]-V$ is transcendental over $\mathbb{R}^V$, and so the translates of $\mathbb{R}^V$ by the powers of $x$ are disjoint. Thus, a Vitali-style argument shows that if $\mathbb{R}^V$ is measurable, it must have measure zero.

Gunter Fuchs and I made this observation in connection with this related MO question, but it may have been known previously.

The answer is no. If $\mathbb{R}^V$ is measurable in a forcing extension $V[G]$ having new reals, then the measure must be $0$. The point is that every new real $x$ in $V[G]$ but not in $V$ is transcendental over $\mathbb{R}^V$, since one cannot add algebraic numbers by forcing. It follows that the translates of $\mathbb{R}^V$ by the powers of $x$ are disjoint. Thus, a Vitali-style argument with wrapped translations of the unit interval of $V$ shows that if $\mathbb{R}^V$ is measurable, it must have measure zero.

Gunter Fuchs and I had made this observation in connection with Sebastian's question Probabilities independent of ZFC?, which is very much related to your question.