jmc

Let's sketch our triangle first. Knowing that the incenter is the intersection of angle bisectors, we draw the angle bisectors as well. Since $\angle BAD=\angle CAD$ by definition and $\angle ABC=\angle ACB$ since $△ABC$ is isosceles, we can see that $\angle ADB=\angle ADC=90^{\circ }.$ Therefore, we see that $AD\perp BC,$ which means that $ID$ is an inradius. What's more, we can find $ID$ using the Pythagorean Theorem, since we have $IC=18$ and $CD=\frac{1}{2}\cdot 30=15.$

Therefore, $ID=\sqrt{I{C}^2-C{D}^2}=\sqrt{18^2-15^2}=\sqrt{99}=\boxed{3\sqrt{11}}.$