We know that existence of a Lebesgue nonmeasurable set is consistent with the Axiom Of Choice. Is the converse true? That is, does the existence of a Lebesgue nonmeasurable set imply the Axiom Of Choice?

No, the existence of a nonLebesgue measurable set does not imply the axiom of choice. If ZF is consistent, then settheorists can construct models of ZF having a nonLebesgue measurable set, but still not satisfying AC. This is quite reasonable, because the existence of a nonLebesgue measurable set is a very local assertion, having to do only with sets of reals, and thus can be satisfied with a small example, by settheoretic standards. The axiom of choice, in contrast, is a global assertion insisting that every set, even a very large set, has a wellorder. So we don't expect to turn a mere nonmeasurable set into wellorderings of enormous sets, such as the power set $P(\mathbb{R})$. And indeed, one can use forcing to produce a model $L(P(\mathbb{R})^{V[G]})$ which satisfies $ZF+\neg AC$, for similar reasons as in the usual $\neg AC$ models, but since it has the true $P(\mathbb{R})$, it will have all the same nonLebesgue measurable sets as in the ambient ZFC universe $V[G]$. (Finally, let me make a minor objection to the question: consistency is a symmetric relation, and so if $A$ is consistent with $B$, then $B$ would be consistent with $A$, and so one wouldn't ordinarily speak of a "converse". You seem instead to be refering to the implication that AC implies there is a nonmeasurable set, and this is how I took your question.) 

