The three medians form the side lengths of a triangle whose area is three-fourths the original area

Use the two known median lengths and that area to find the two possible included angles, then reject the case that makes two medians equal

The triangle whose sides are the three medians has area $$\frac34\left(3\sqrt{15}\right)=\frac{9\sqrt{15}}4.$$ If $\theta$ is the included angle between its sides $3$ and $6,$ then $$9\sin\theta=\frac{9\sqrt{15}}4,$$ so $\cos\theta=\pm\frac{1}{4}.$ By the law of cosines, the third median $m$ satisfies $$m^2=3^2+6^2-2(3)(6)\cos\theta,$$ giving $m^2=36$ or $54.$ The value $m=6$ would make two medians, and hence two sides, equal. Because the triangle has unequal sides, $m=\sqrt{54}=3\sqrt6.$

Therefore, the correct answer is C.