Answer
The ratio of the charge density on sheet 3 to that on sheet 2 is $~~-1.5$
Work Step by Step
Between sheet 2 and sheet 3, the net electric field is $E_s$
To the right of sheet 3, the net electric field is $0$
Then the magnitude of the electric field due to $\sigma_3$ must be $\frac{E_s}{2}$ and $\sigma_3$ must be a negative charge density.
Between sheet 2 and sheet 3, the net electric field is $E_s$
Between sheet 1 and sheet 2, the net electric field is $\frac{E_s}{3}$
Then the magnitude of the electric field due to $\sigma_2$ must be $\frac{E_s}{3}$ and $\sigma_2$ must be a positive charge density.
Note that the electric field is proportional to the surface charge density.
We can find the ratio $\frac{\sigma_3}{\sigma_2}$:
$\frac{\sigma_3}{\sigma_2} = -\frac{\frac{E_s}{2}}{\frac{E_s}{3}} = -1.5$
The ratio of the charge density on sheet 3 to that on sheet 2 is $~~-1.5$
