Explanation

(b) Let nucleus be $\ _{Z}X^{A}$. Nuclear radius, $R = R_{0}A^{1/3}$ where $R_{0}$ is a constant whose value is found to be $1.2 \times 10^{- 15}m$ and $A$ is the mass number
$\therefore\frac{R_{X}}{R_{Cs}} = \left( \frac{A}{189} \right)^{1/3},\therefore\frac{1}{3} = \left( \frac{A}{189} \right)^{1/3}$
$A = \frac{189}{3^{3}} = \frac{189}{27} = 7$
The given nucleus is $Li^{7}$