Correct option is (a) 66%
To explain I would say: In DSB-SC carrier is suppressed.
So Total Power required in DSBSC Modulation = (u^2XPc)/2 = Pc/2
In normal AM, carrier is not suppressed.
So total power required in AM Modulation = (1+(u^2/2))XPc = 3Pc/2
Therefore, Power saving = ((Pc/2)/(3Pc/2)) x 100% = 66%.