Let the length, width, and height of the prism be $s_{1}, s_{2}, s_{3}$. Without loss of generality, assume that $s_{1} \leq s_{2} \leq s_{3}$. Then, we have that $s_{1} s_{2}=48$ and $s_{2} s_{3}=240$. Noting that $s_{1} \leq s_{2}$, we must have $\left(s_{1}, s_{2}\right)=(1,48),(2,24),(3,16),(4,12),(6,8)$. We must also have $s_{2} s_{3}=240$ and $s_{2} \leq s_{3}$, and the only possibilities for $\left(s_{1}, s_{2}\right)$ that yield integral $s_{3}$ that satisfy these conditions are $(4,12)$, which gives $s_{3}=20$, and $(6,8)$, which gives $s_{3}=30$. Thus, the only valid $\left(s_{1}, s_{2}, s_{3}\right)$ are $(4,12,20)$ and $(6,8,30)$. It follows that the only possible areas of the third face are $4(20)=80$ and $6(30)=180$, so the desired answer is $80+180=260$.