If the atom of carbon has lost 4 electrons, it has now an excess of charge equal to +4e:
[tex]Q=+4e[/tex]
where [tex]e=1.6 \cdot 10^{-19}C[/tex]
Its charge is concentrated in the nucleus, so we can treat it as a single-point charge, whose electric field is given by:
[tex]E=k \frac{Q}{r^2} [/tex]
where k is the Coulomb's constant and r is the distance from the charge. In our problem,
[tex]r=16 nm=16 \cdot 10^{-9} m[/tex]
therefore the electric field at this distance is
[tex]E=k \frac{Q}{r^2}=(8.99 \cdot 10^{9} Nm^2C^{-2}) \frac{4(1.6 \cdot 10^{-19}C)}{(16\cdot 10^{-9}m)^2}= 2.25 \cdot 10^7 N/C[/tex]
