I take it your parameters are set up so $k=3$ and $\lambda=1$. In addition to $bk=vr$, there is another relation that holds among the parameters. Do you know what relation I'm referring to? If not, you can find it in the wikipedia article on BIBDs. Then see whether the two relations, along with $k=3$ and $\lambda=1$, aren't enough to get the result on $v$.
EDIT: Note that BIBD with $k=3$, $\lambda=1$ are called "Steiner triple systems," and there is a lot of information about them available on the web and in texts.
FURTHER EDIT: First, use those two relations to get $r=(v-1)/2$ and $b=v(v-1)/6$. Then, note that $r$ is an integer, so $v$ must be odd. So, what are the possible remainders when you divide $v$ (which is odd) by 6? And what gets ruled out by knowing that $b$ has to be an integer?