Each infinite sheet of density $\sigma$ creates a uniform field of magnitude $\sigma /(2{ϵ}_0)$ pointing away (toward) the sheet for positive (negative) $\sigma$. Take the rightward direction as positive and sum contributions from all six sheets in each region.

Configuration I with densities $(+{\sigma }_0,-{\sigma }_0,+{\sigma }_0,-{\sigma }_0,+{\sigma }_0,-{\sigma }_0)$:

$E_{\text{region 4}}=\frac{{\sigma }_0}{2{ϵ}_0}[1-1+1-1-1+1]=0$. (A) is correct.

Potential drop from sheet 1 to sheet 6, with $d=1\mu$m between consecutive sheets, using fields in regions 1 through 5:

$E_1,E_2,E_3,E_4,E_5=\frac{{\sigma }_0}{2{ϵ}_0}\{-1,+1,-1,0,-1\}\cdot (\text{signs by sheet contributions})$. Working it out with the contribution method:

$V_{\text{first}}-V_{\text{last}}=\frac{3{\sigma }_{0}d}{{ϵ}_0}=\frac{3\times 9\times 10^{-6}\times 10^{-6}}{9\times 10^{-12}}=3$ V $\ne 5$ V. (C) is incorrect.

Configuration II with densities $(+{\sigma }_0/2,-{\sigma }_0,+{\sigma }_0,-{\sigma }_0,+{\sigma }_0,-{\sigma }_0/2)$:

$E_{\text{region 3}}=\frac{{\sigma }_0}{2{ϵ}_0}\left[\frac{1}{2}-1+1-1-1+\frac{1}{2}\right]=-\frac{{\sigma }_0}{2{ϵ}_0}$

So the magnitude is ${\sigma }_0/(2{ϵ}_0)$, not ${\sigma }_0/{ϵ}_0$. (B) is incorrect.

Computing $V_{\text{first}}-V_{\text{last}}$ in configuration II using the field summary gives $\frac{{\sigma }_{0}d}{2{ϵ}_0}\ne 0$. (D) is incorrect.

$\therefore$ Only (A) is correct.