Ratio of the power in the sidebands to the total power in conventional AM Message signal and carrier signal are given as 

$m=\cos(2 \pi  \cdot f_{m} \cdot t) $

$c=A_{c}\cdot \cos(2 \pi  \cdot f_{c} \cdot t) $

I was asked to find modulation index $a$ so that ratio of the power in the sidebands to the total power in the transmitted signal will be $\frac{1}{3}$

My attempt

We know that, in conventional AM, total power of transmitted signal $P_{u}= \frac{Ac^2}{2} + \frac{Ac^2}{2}\cdot a^2 P_{m} $  where $P_{m} $ is message signal's power.

I think that power in sidebands is related to $\frac{Ac^2}{2}\cdot a^2 P_{m}$ term

$\frac{Ac^2}{2}\cdot a^2 P_{m} = \frac{1}{3}P_{u}$

Submitting $P_{u}$

$\frac{Ac^2}{2}\cdot a^2 P_{m} = \frac{1}{3}(\frac{Ac^2}{2}(1 +  a^2 P_{m}))$

$ a^2 P_{m} = \frac{1}{3}(1 +  a^2 P_{m})$

$ a^2 =  \frac{1}{2P_{m}}$

It seems false because modulation index might have been a numerical value.
this is John BG jgb2012@sky.com

In AM modulation, as shown in

https://en.wikipedia.org/wiki/Amplitude_modulation#Modulation_index

the modulation index is the ratio between the amplitude of the modulated and the amplitude of the carrier.

So the sought modulation index $a=\frac{1}{Ac}$

modulating signal power: $\frac{1}{2}$

carrier power: $\frac{Ac^2}{2}$

side bands to total power ratio:   $\frac{1}{(1+\frac{Ac^2}{2})}$

solving:  $\frac{1}{(1+\frac{Ac^2}{2})} =\frac{1}{3}$ gives $Ac=2$

so the answer is $a=\frac{1}{2}$