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 15 = P_c \big ( 1 + \large\frac{0.16^2}{2}$$\big )$

$\Rightarrow P_c = \large\frac{15}{1.013}$$ = 14.81\;W$

The total power in both side bands $ = 15-14.81 = 0.19\;W$

The power in each side band is half of this $ \approx 0.095\;W$

$\Rightarrow$ The fraction of the power in the carrier is $\large\frac{14.81}{15} $$\approx 0.99$