Answer
(a) $7.0fm$
(b) $4.7fm$
Work Step by Step
(a) We can calculate the nuclear radius of $Au^{197}_{79}$ as
$r=(1.2\times 10^{-15}m)A^{\frac{1}{3}}$
$\implies r=(1.2fm)(197)^{\frac{1}{3}}$
$r=7.0fm$
(b) We can calculate the nuclear radius of $Co^{60}_{27}$ as
$r=(1.2\times 10^{-15}m)A^{\frac{1}{3}}$
$\implies r=(1.2fm)(60)^{\frac{1}{3}}$
$r=4.7fm$
