jmc

The volume is the pyramid is $\frac{1}{3}{s}^{2}h$, where $s$ is the side length of the base and $h$ is the height of the pyramid. Therefore, the area of the base is $s^2=(63,960{\text{ m}}^3)/\left(\frac{1}{3}\cdot 30\text{ m}\right)=6396$ square meters. Calling the center of the base $D$, we apply the Pythagorean theorem to triangle $ABD$ to get $$AB=\sqrt{h^2+(s/2{)}^2}=\sqrt{h^2+s^2/4}=\sqrt{30^2+6396/4}=\sqrt{2499},$$ which is closer to $\sqrt{2500}=\boxed{50}$ meters than to $\sqrt{2401}=49$ meters, since $49.5^2=2450.25$.