Asingle gram of 235u fuel can heat and boil 2.49 x 107 g of water. what volume of water is this?

You just have to convert the mass of water into volume.

To do that you use the density of water, which is about 1.0 g/ ml

So, from the formula of density D = M / V, you get V = M / D

=> V = 2.49 * 10^7 grams / 1.0 g / ml = 2.49 * 10 ^ 7 ml

You can pass that to liters using the conversion factor 1000 ml = 1 l

2.49 * 10^7 ml * 1 l / 1000 ml = 2.49 * 10^4 l = 24,900 l

24,900 l

Answer is to avoid experimental error