smc

Let $△ABC$ be an isosceles triangle with $AB=AC=12,$ and $BC=6$. Draw altitude $\overline{AD}$. Since $A$ is the apex of the triangle, the altitude $\overline{AD}$ is also a median of the triangle. Therefore, $BD=CD=3$. Using the Pythagorean Theorem, $AD=\sqrt{12^2-3^2}=3\sqrt{15}$. The area of $△ABC$ is $\frac{1}{2}\cdot 6\cdot 3\sqrt{15}=9\sqrt{15}$. If $a,b,c$ are the sides of a triangle, and $A$ is its area, the circumradius of that triangle is $\frac{abc}{4A}$. Using this formula, we find the circumradius of $△ABC$ to be $\frac{6\cdot 12\cdot 12}{4\cdot 9\sqrt{15}}=\frac{2\cdot 12}{\sqrt{15}}=\frac{24\sqrt{15}}{15}=\frac{8\sqrt{15}}{5}$. The answer is $\boxed{\text{none of these}}$