Answer
$E=2\times 10^{-11}\frac{N}{C}$
Work Step by Step
We know that:
$E=\frac{\sigma}{2\epsilon_{\circ}}$
We plug in the known values to obtain:
$E=\frac{1.77\times 10^{-22}}{2(8.85\times 10^{-12})}=1\times 10^{-11}$
This is the electric field due to one sheet. Thus, the electric field due to two sheets is twice as that of one sheet;
$\implies E=2(1\times 10^{-11})=2\times 10^{-11}\frac{N}{C}$
