I think the right solution is c). I'll pass on my reasoning to you:
R=6\, \textrm{cm}=0'06\, \textrm{m}
\sigma =\dfrac{10}{\pi} \, \textrm{nC/m}^2=\dfrac{1\cdot 10^{-8}}{\pi}\, \textrm{C/m}^2
P=0'03\, \textrm{m}
P'=10\, \textrm{cm}=0,1\, \textrm{m}
Point P:
\left.
\phi =\oint E\cdot...