The ratio of the Nuclear density of the gold isotope ${ }_{79}^{197} Au$ and the silver isotope ${ }_{47}^{107} Ag$ is:

$\left(\frac{197}{107}\right)^{\frac{1}{3}}$ $\left(\frac{79}{47}\right)^{\frac{1}{3}}$ $\left(\frac{197}{107}\right)^{\frac{2}{3}}$ $1: 1$

$1: 1$

The correct answer is Option (4) → $1: 1$