Answer :
The electric field between the plates of a capacitor (assuming it is closely spaced) is given by:
[tex]\vec{E}=\frac{Q}{\epsilon_0A}[/tex]We can replace our values, and we'll get the following:
[tex]926331.13=\frac{1675802144.45*1.6*10^{-19}}{8.8541878*10^{-12}A}[/tex]By isolating the are, we get:
[tex]A=\frac{1675802144.45*1.6*10^{-19}}{8.8541878*10^{-12}*926331.13}=3.269*10^{-5}m^2[/tex]Now, if we replace it on the area of a circle:
[tex]3.269*10^{-5}=\pi r^2[/tex]Our value of r is:
[tex]r=\sqrt[\placeholder{⬚}]{\frac{3.269*10^{-5}}{\pi}}=3.2258mm[/tex]Then, our final answer is d=6.4515mm