The correct answer is $\boxed{\text{B) }V}$.
The triple integral $\iint\limits_{\text{S}} {\overrightarrow {\text{r}} \cdot {\text{d}}\overrightarrow {\text{S}} }$ is the surface integral of the position vector over a closed surface $S$. It can be interpreted as the volume enclosed by $S$.
To see this, consider a small surface element $d\overrightarrow {\text{S}}$ on $S$. The vector $d\overrightarrow {\text{S}}$ is a vector normal to $S$, and its magnitude is the area of $d\overrightarrow {\text{S}}$. The dot product of $\overrightarrow {\text{r}}$ and $d\overrightarrow {\text{S}}$ is the projection of $\overrightarrow {\text{r}}$ onto $d\overrightarrow {\text{S}}$. Therefore, the surface integral $\iint\limits_{\text{S}} {\overrightarrow {\text{r}} \cdot {\text{d}}\overrightarrow {\text{S}} }$ is the sum of the projections of $\overrightarrow {\text{r}}$ onto all the surface elements $d\overrightarrow {\text{S}}$ on $S$.
If $S$ is a closed surface that encloses volume $V$, then the projections of $\overrightarrow {\text{r}}$ onto all the surface elements $d\overrightarrow {\text{S}}$ on $S$ will add up to the volume $V$. Therefore, the surface integral $\iint\limits_{\text{S}} {\overrightarrow {\text{r}} \cdot {\text{d}}\overrightarrow {\text{S}} }$ is equal to the volume $V$.