The total power in an AM signal when the carrier power and the percentage of modulation are known is $P_T = P_c \big ( 1 + \large\frac{m^2}{2}$$\big )$

$\Rightarrow 32 = P_c \big ( 1 + \large\frac{0.28^2}{2}$$\big )$

$\Rightarrow P_c = \large\frac{32}{1.0392}$$ = 30.793\;W$

The total power in both side bands $ = 32-30.793 = 1.207\;W$

The power in each side band is half of this $ \approx 0.60\;W$

$\Rightarrow$ The fraction of the power in the carrier is $\large\frac{30.79}{32} $$\approx 0.96$