The electron charge is equal to [tex]-e=-1.6\cdot 10^{-19}C[/tex]. The atomic nucleus of the problem has a charge of [tex]+40 e=40\cdot (1.6\cdot 10^{-19}C)=6.4\cdot 10^{-18}C[/tex]. The distance between the nucleus and the electron is [tex]r=10^{-9}m[/tex], so we can calculate the electrostatic (Coulomb) force between the two:
[tex]F=k_e \frac{(-e)(+40e) }{r^2} =8.99\cdot 10^9 Nm^2C^{-2} \frac{(-1.6\cdot 10^{-19}C)(6.4\cdot 10^{-18}C)}{(10^{-9}m)^2} =[/tex]
[tex]=-9.2 \cdot 10^{-9} N[/tex]
which is attractive, since the two charges have opposite sign.
