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 25 = P_c \big ( 1 + \large\frac{0.3^2}{2}$$\big )$

$\Rightarrow P_c = \large\frac{25}{1.045}$$ = 23.92\;W$

The total power in both side bands $ = 25 - 23.92 = 1.08\;W$

The power in each side band is half of this $ = 0.54\;W$

$\Rightarrow$ The fraction of the power in the carrier is $\large\frac{23.92}{25} $$= 0.957$