jmc

We know the rectangular base of the pyramid has area $48$. To find the volume, we must also determine the height.

Let the rectangular base be $ABCD$. Let the apex of the pyramid be $X$, and let $O$ be the foot of the perpendicular drawn from $X$ to face $ABCD$: Then by the Pythagorean theorem, we have $$\begin{array}{rl}13^2& =O{X}^2+O{A}^2=O{X}^2+O{B}^2\\ & =O{X}^2+O{C}^2=O{X}^2+O{D}^2.\end{array}$$Therefore, $OA=OB=OC=OD$, so $O$ must be the center of the rectangle (where the perpendicular bisectors of the sides meet). This is also the point where the diagonals of $ABCD$ bisect each other. Each diagonal of $ABCD$ has length $\sqrt{6^2+8^2}=10$, so we have $OA=OB=OC=OD=5$.

Thus $OX=\sqrt{13^2-O{D}^2}=\sqrt{13^2-5^2}=12$, and so the height of the pyramid is $12$. The volume is $$\begin{array}{rl}\frac{1}{3}\cdot (\text{area of base})\cdot (\text{height})& =\frac{1}{3}\cdot 48\cdot 12\\ & =16\cdot 12\\ & =\boxed{192}.\end{array}$$