Answer
The net charge on the shell is $~~7.0\times 10^{-6}~C$
Work Step by Step
At $r = 2.0~cm$, the electric field is $3\times 10^7~N/C$
We can find the charge of the particle at the center:
$E = \frac{q_p}{4\pi~\epsilon_0~r^2}$
$q_p = E~4\pi~\epsilon_0~r^2$
$q_p = (3\times 10^7~N/C)~(4\pi)~(8.854\times 10^{-12}~F/m)~(0.020~m)^2$
$q_p = 1.335\times 10^{-6}~C$
At $r = 5.0~cm$, the electric field is $3\times 10^7~N/C$
We can find the total charge of the particle at the center plus the charge on the shell:
$E = \frac{q}{4\pi~\epsilon_0~r^2}$
$q = E~4\pi~\epsilon_0~r^2$
$q = (3\times 10^7~N/C)~(4\pi)~(8.854\times 10^{-12}~F/m)~(0.050~m)^2$
$q = 8.345\times 10^{-6}~C$
We can find the net charge on the shell:
$(8.345\times 10^{-6}~C) -(1.335\times 10^{-6}~C) = 7.0\times 10^{-6}~C$
The net charge on the shell is $~~7.0\times 10^{-6}~C$
