Answer: C

Small Hint:

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

Big Hint:

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

Solution:

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.