Here we go... I hope this makes sense without diagrams:
The table that was provided is a unit-hydrograph. The CFS values are based on a per inch of rainfall (cfs/in) for a one-hour rainfall intensity.
Since the problem states that a 2-hour rainfall had occurred, then you need to add up the Q/in values (0+75+200+100+50+25+0=450) and multiply by 2 for the unit CFS (900 cfs/in) and multiply again by 0.5" for the actual CFS (450 cfs).
However, to properly calculate the total volume of water, we need to graph the actual CFS at each time interval and calculate the area below the line.
Here are the data domains to develop the volume by graph:
T (hours) = {0,1,2,3,4,5,6}
Qu = {0,75,200,100,50,25,0}
Q1 = {0,37.5,100,50,25,12.5,0}
Q2 = {0,0,37.5,100,50,25,12.5} (Q2 = Q1 shifted 1 hour)
QTOT = {0,37.5,137.5,150,75,37.5,12.5}
If you graph QTOT (cfs) vs. T (hr), the area under the line = 450 cfsh (units are cfs x h = volume of water).
Therefore 450 cfsh = 450 cfsh x 3600 sec/hr x 1hr / 43560 cf/AF = 37.2 AF
I think if you do a few of these problems, you might find that volume of water is the same result as the sum of QTOT, but I can't verify at this time.
Good luck!