Surface charge density is

$$\rho = {Q \over {4\pi {R^2}}}$$

where R is the radius of the sphere. For cancelling the field inside the cavity due to the shell of radius R, the shell with radius 2R induces a charge $$-$$Q in its inner surface and hence the total charge on the outer surface is $$Q_1+Q_2$$. Similarly, the charge on the outermost shell is

$${Q_1} + {Q_2} + {Q_3}$$

It is given that the surface charge densities $$\rho_1,\rho_2$$ and $$\rho_3$$ are equal, that is,

$${{{Q_1}} \over {4\pi {R^2}}} = {{{Q_1} + {Q_2}} \over {4\pi {{(2R)}^2}}} = {{{Q_1} + {Q_2} + {Q_3}} \over {4\pi {{(3R)}^2}}}$$

$$ \Rightarrow {Q_1} = {{{Q_1} + {Q_2}} \over 4} = {{{Q_1} + {Q_2} + {Q_3}} \over 9}$$

That is,

$${Q_1} = {{{Q_1}} \over 4} + {{{Q_2}} \over 4}$$

$$ \Rightarrow {Q_2} = 3{Q_1}$$

Therefore,

$${Q_3} = 5{Q_1}$$

Hence, the ratio of the charges given to the shells $${Q_1}:{Q_2}:{Q_3}$$ is 1 : 3 : 5.