Here the question is about "dependency" not hazard ..So we see by Bernstein's conditions..So for RAW dependency , we need to check that write of an earlier instruction is succeeded by some subsequent instruction and following this we have 4 dependencies
a) I1 - I2
b) I1 - I3
c) I2 - I3
d) I2 - I4..
So in all we have 4 RAW dependencies here..So correct answer..Remember that hazard is not same as dependency..Hazard finding needs other factors also like number of stages etc..