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 $V[G]-V$ V[G]$but not in$V$is transcendental over$\mathbb{R}^V$, and so since one cannot add algebraic numbers by forcing. It follows that the translates of$\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$is measurable, it must have measure zero. Gunter Fuchs and I had made this observation in connection with this related MO Sebastian's question Probabilities independent of ZFC?, but it may have been known previouslywhich is very much related to your question. 1 The answer is no. If$\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.