Answer: 721

Solution:

For a box with dimensions $a, b, c,$ the conditions are $2(ab + bc + ca) = 54$ and $abc = 23,$ so $ab + bc + ca = 27.$ The smallest sphere containing a box has the box's space diagonal as a diameter, so $$\begin{aligned} &r^2 = \max_{\mathcal{B}} \frac{a^2 + b^2 + c^2}{4} \\ &= \max_{\mathcal{B}} \frac{(a + b + c)^2 - 54}{4}. \end{aligned}$$

With $ab + bc + ca$ and $abc$ fixed, $s = a + b + c$ ranges over an interval, and at an endpoint the cubic $t^3 - st^2 + 27t - 23$ has a double root, meaning two dimensions coincide. Setting $b = c:$ $2ab + b^2 = 27$ and $ab^2 = 23,$ so eliminating $a$ gives $\frac{b(27 - b^2)}{2} = 23,$ i.e. $b^3 - 27b + 46 = 0,$ which factors as $(b - 2)(b^2 + 2b - 23) = 0.$ The roots are $b = 2$ and $b = 2\sqrt{6} - 1.$

For $b = 2,$ $a = \frac{23}{4}$ and $s = \frac{23}{4} + 4 = \frac{39}{4} = 9.75;$ for $b = 2\sqrt{6} - 1,$ $s \approx 9.31$ is smaller. So the maximum of $a^2 + b^2 + c^2$ is $\left(\frac{39}{4}\right)^2 - 54 = \frac{657}{16},$ giving $r^2 = \frac{657}{64}$ and $p + q = 657 + 64 = 721.$