The extent of ionisation in a 0.75M solution of a monoprotic acid is 0.040. calculate Ka
0.75 x 0.04 = 0.03
0.75-0.03 = 0.72
........HA ==> H^+ + A^-
I.....0.75.....0.....0
C....-0.03....0.03...0.03
E....0.72......0.03...0.03
Substitute the E line into the Ka expression and solve for Ka.
I assume that is 0.04 ad not 0.04%.