Is there enough space in the box to fit one more battery?

If diameter of a battery is 1, then distance between rows of batteries (center to center) is the same as height of the equilateral triangle with side 1, or $\sqrt{1 - \left(\frac{1}{2}\right)^2}=\frac{\sqrt{3}}{2}$. Total height of 9 rows is $0.5 + 8 \frac{\sqrt{3}}{2} + 0.5 \approx 7.93$, less than 8.