Answer
a. 5.8 fm
b. $\approx$ 29
Work Step by Step
Use equation 30–1.
a.
$$r=(1.2\times10^{-15}m)A^{1/3}=(1.2\times10^{-15}m)112^{1/3}=5.8 fm$$
b.
$$r=(1.2\times10^{-15}m)A^{1/3} $$
$$A=(\frac{r}{1.2\times10^{-15}m })^{3} $$
$$A=(\frac{3.7\times10^{-15}m }{1.2\times10^{-15}m })^{3}\approx 29 $$