Answer
$2.7 \times 10^{-3}N$
Work Step by Step
According to $F=k\frac{Q_{1}Q_{2}}{r^{2}}$
The charge of the iron nucleus is 26e and the charge of one electron is e, plug these two to $Q_{1}$ and $Q_{2}$.
The distance between them is $1.5 \times 10^{-12}$, plug it to r.
Then plug in k, the constant $9.0 \times 10^{9} N \times m^{2}/C^{2}$.
It's $F=9.0 \times 10^{9} N \times m^{2}/C^{2} \times \frac{26e \times e}{2.25 \times 10^{-24}}$
Because $e = 1.6^{-19} C$
$F = 9.0 \times 10^{9} N \times m^{2}/C^{2} \times \frac{26 \times 1.6^{-19} C \times 1.6^{-19} C}{2.25 \times 10^{-24}}$ = $2.7 \times 10^{-3}N$
