Pinging Protons

An iron nucleus has 26 protons each having a charge of $1.6 \times 10^{-19}$ , which is the charge on one proton.Using Coulomb's law force on the proton = $(kq_1q_2)/r^2$ where r is the distance between the charges.Here $q_1$ = 26×1.6×10^(-19) and. $q_2$ = 1.6×10^(-19) and. r=5×10^(-15). Plugging in the values you get force=239.616 N.

Formula for coulombic force:

$k\frac{q_1 q_2}{d^2}$

where $d$ is the distance of seperation $= (2+3) fm = 5 fm$

$q_1$ is the charge of the pinging proton $= 1.60218 \times10^{-19} C$

$q_2$ is the charge of the nucleus of iron with 26 protons $= 26 \times 1.60218 \times10^{-19} C$

You get $k = 9 \times 10^9$ from the question.

So

$\frac{\left(9 \times 10^9\right) (26 \times 1.60218 \times 1.60218)}{10^{19} \times 10^{19} \left(\frac{5}{10^{15}}\right)^2} = \boxed{240.268}$ is the answer in Newtons

Use Coulumb's Law using $1.60 \times 10^{-19} C$ as the charge of proton and $26(1.60 \times 10^{-19}) C$ as the charge of the iron nucleus and its distance will be the distance between the proton and the center of the iron nucleus which is $5 fm$ or $5 \times 10^{-15} m$