The mass number of nucleus having radius equal to half of the...

Given: ${R}_{1} = \dfrac{{R}_{2}}{2}$

As we know, size of the nucleus is given by $R = {R}_{0}{\left(A\right)}^{1 / 3}$

Therefore, we can write ${R}_{0}{\left({A}_{1}\right)}^{1 / 3} = \dfrac{{R}_{0}}{2}{\left({A}_{2}\right)}^{1 / 3}$

$⇒ {A}_{1} = \dfrac{1}{8}{A}_{2}$

$⇒ {A}_{1} = \dfrac{192}{8} = 24$