Answer
$E = 5.3\times 10^7~N/C$
Work Step by Step
We can think of the charged metal plate as a slab of charge and we can find $\sigma$:
$\sigma = \frac{6.0\times 10^{-6}~C}{(0.080~m)^2} = 9.375\times 10^{-4}~C/m^2$
We can find the magnitude of the electric field:
$E = \frac{\sigma}{2\epsilon_0}$
$E = \frac{9.375\times 10^{-4}~C/m^2}{(2)(8.854\times 10^{-12}~F/m)}$
$E = 5.3\times 10^7~N/C$
