first off, let's change the mixed fractions to improper fractions and let's check what the volume in in³ for the prism is.
[tex]\stackrel{mixed}{2\frac{2}{5}}\implies \cfrac{2\cdot 5+2}{5}\implies \stackrel{improper}{\cfrac{12}{5}} ~\hfill \stackrel{mixed}{3\frac{1}{5}}\implies \cfrac{3\cdot 5+1}{5}\implies \stackrel{improper}{\cfrac{16}{5}} \\\\\\ \stackrel{\textit{volume of the prism}}{ \cfrac{12}{5}\cdot \cfrac{16}{5}\cdot 2\implies \cfrac{384}{25}}~in^3[/tex]
now, we know the small cubes have an "edge" or namely a side of 1/5, let's get their volume as well [tex]\cfrac{1}{5}\cdot \cfrac{1}{5}\cdot \cfrac{1}{5}\implies \cfrac{1}{125}~in^3[/tex]
now, how many times does the volume of one small cube, go into the volume of the containing prism?
[tex]\stackrel{\textit{\large volumes}}{\cfrac{~~~~ \stackrel{prism}{\frac{384}{25}}~~~~}{\underset{cube}{\frac{1}{125}}}}\implies \cfrac{384}{25}\cdot \cfrac{125}{1}\implies \boxed{1920}[/tex]
