If $G$ has a homomorphism onto a finite cyclic group, then it is inner amenable? (For every subset count the number of elements in the factor.) If so then the free Burnside group is certainly inner amenable. On the other hand, by Zelmanov's theorem, there exists a finite index subgroup of $B(2,665)$ which does not have finite factors. That group also is i.c.c., etc. So you may want to ask your question about that group. What is the reason for this question? There is a "similar" and quite popular question of whether $B(2,n)$, $n\gg 1$, has property (T). Y. Shalom conjectured in his ICM talk that it has. Gromov (unpublished) conjectured that it has not.
Edit: I was wrong in the first statement. The counting does not produce a measure because two disjoint subsets can map to the same set in the factor-group. So the question about inner amenability of $B(2,665)$ is open.
If $G$ has a homomorphism onto a finite cyclic group, then it is inner amenable? (For every subset count the number of elements in the factor.) If so then the free Burnside group is certainly inner amenable. On the other hand, by Zelmanov's theorem, there exists a finite index subgroup of $B(2,665)$ which does not have finite factors. That group also is i.c.c., etc. So you may want to ask your question about that group. What is the reason for this question? There is a "similar" and quite popular question of whether $B(2,n)$, $n\gg 1$, has property (T). Y. Shalom conjectured in his ICM talk that it has. Gromov (unpublished) conjectured that it has not.