The total power content of AM wave is 1320 W. The percentage modulation if each side band contains 160 W is 

$$\begin{gathered} P_T=1320 W \\ We know, \frac{1}{2}{P}_{SB}=P_{LSB } and \frac{1}{2}{P}_{SB}=P_{USB} \\ \therefore P_{SB}=2P_{LSB}=2\times 160 = 320 W \\ Also , P_T=P_c+P_{SB} \\ \Rightarrow 1320=P_c+320 \\ \Rightarrow P_c=1000 W \\ Now, \frac{P_T}{P_c}=1+\frac{m^2}{2} \\ \Rightarrow \frac{1320}{1000}=1+\frac{m^2}{2} \\ \Rightarrow 1.32=1+\frac{m^2}{2} \\ \Rightarrow 0.32=\frac{m^2}{2} \\ \Rightarrow m^2 = 0.64 \\ \Rightarrow m=0.8 \\ i.em\%=m\times 100=0.8\times 100= 80\% \end{gathered}$$