Electrostatics

Surface charge density is
ρ =

Q

4πR2

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 surfce and hence the total charge on the outer surface is Q1+Q2. Simillarly, the charge on the outermost shell is
Q1+Q2+Q3

It is given that the surface charge densities ρ1,ρ2 and ρ3 are equal, that is,

Q1

4πR2

=

Q1+Q2

4π(2R)2

=

Q1+Q2+Q3

4π(3R)2

⇒ Q1 =

Q1+Q2

4

=

Q1+Q2+Q3

9

That is,
Q1 =

Q1

Q4

+

Q2

4

⇒ Q2 = 3Q1
Therefore,
Q3 = 5Q1
Hence, the ratio of the charges given to the shells Q1:Q2:Q3 is 1 : 3 : 5.