Answer
The net charge enclosed by the outer surface is $~~+2.0\times 10^{-9}~C$
Work Step by Step
We can find magnitude of the net charge $q$ enclosed by the outer surface:
$\epsilon_0~\Phi = q$
$\epsilon_0~E~4\pi~r^2 = q$
$q = (8.854\times 10^{-12}~F/m)(450~N/C)(4\pi)(0.20~m)^2$
$q = 2.0\times 10^{-9}~C$
Since the electric field is directed outward, the net charge enclosed by the outer surface is $~~+2.0\times 10^{-9}~C$
