jmc

We look at $\angle ABC$. $\angle ABC$ cuts off minor arc $\widehat{AC}$, which has measure $2\cdot m\angle ABC=150^{\circ }$, so minor arcs $\widehat{AB}$ and $\widehat{BC}$ each have measure $\frac{360^{\circ }-150^{\circ }}{2}=105^{\circ }$.

Stuart cuts off one $105^{\circ }$ minor arc with each segment he draws. By the time Stuart comes all the way around to his starting point and has drawn, say, $n$ segments, he will have created $n$ $105^{\circ }$ minor arcs which can be pieced together to form a whole number of full circles, say, $m$ circles. Let there be $m$ full circles with total arc measure $360^{\circ }\cdot m$. Then we have $$105^{\circ }\cdot n=360^{\circ }\cdot m.$$ We want to find the smallest integer $n$ for which there is an integer solution $m$. Dividing both sides of the equation by $15^{\circ }$ gives $7n=24m$; thus, we see $n=24$ works (in which case $m=7$). The answer is $\boxed{24}$ segments.