jmc

As we are dealing with volumes, the ratio of the volume of $P^{\prime }$ to $P$ is the cube of the ratio of the height of $P^{\prime }$ to $P$. Thus, the height of $P$ is $\sqrt[3]{8}=2$ times the height of $P^{\prime }$, and thus the height of each is $12$. Thus, the top of the frustum is a rectangle $A^{\prime }{B}^{\prime }{C}^{\prime }{D}^{\prime }$ with $A^{\prime }{B}^{\prime }=6$ and $B^{\prime }{C}^{\prime }=8$. Now, consider the plane that contains diagonal $AC$ as well as the altitude of $P$. Taking the cross section of the frustum along this plane gives the trapezoid $AC{C}^{\prime }{A}^{\prime }$, inscribed in an equatorial circular section of the sphere. It suffices to consider this circle. First, we want the length of $AC$. This is given by the Pythagorean Theorem over triangle $ABC$ to be $20$. Thus, $A^{\prime }{C}^{\prime }=10$. Since the height of this trapezoid is $12$, and $AC$ extends a distance of $5$ on either direction of $A^{\prime }{C}^{\prime }$, we can use a 5-12-13 triangle to determine that $A{A}^{\prime }=C{C}^{\prime }=13$. Now, we wish to find a point equidistant from $A$, $A^{\prime }$, and $C$. By symmetry, this point, namely $X$, must lie on the perpendicular bisector of $AC$. Let $X$ be $h$ units from $A^{\prime }{C}^{\prime }$ in $AC{C}^{\prime }{A}^{\prime }$. By the Pythagorean Theorem twice,$$\begin{array}{rl}{5}^2+h^2& =r^2\\ 10^2+(12-h{)}^2& =r^2\end{array}$$Subtracting gives $75+144-24h=0⟹h=\frac{73}{8}$. Thus $XT=h+12=\frac{169}{8}$ and $m+n=\boxed{177}$.