Answer
The shell's linear charge density is $~~\lambda = -5.8\times 10^{-9}~C/m$
Work Step by Step
We can draw a Gaussian cylinder of radius $r = 3.5~cm$ and length $L = 1.0~m$ that has the same axis as the charged cylinders.
When $r$ is slightly less than $3.5~cm$, the electric field is due to the charge on the smaller cylinder. When $r$ is slightly more than $3.5~cm$, then the electric field is due to both charged cylinders.
From the graph, we can see that the electric field at $r = 3.5~cm$ due to the shell is $E = -2000~N/C-1000~N/C = -3000~N/C$
We can find the charge $q$ on the shell:
$\epsilon_0~\Phi = q$
$q = \epsilon_0~E~2\pi~r~L$
$q = (8.854\times 10^{-12}~F/m)~(-3000~N/C)~(2\pi)~(0.035~m)~(1.0~m)$
$q = -5.8\times 10^{-9}~C$
Since the length is $L = 1.0~m$, the shell's linear charge density is $~~\lambda = -5.8\times 10^{-9}~C/m$
