Answer
a) $R=180m$
b) $3.5\times10^4$
Work Step by Step
a) $\rho_{nuc}=\frac{m_{nuc}}{V_{nuc}}=\frac{1.6605\times10^{-27}kg}{\frac{4}{3}\pi(1.2\times10^{-15})^3}=2.294\times10^{17}\frac{kg}{m^3}$
$R=\bigg(\frac{M}{\frac{4}{3}\pi\rho_{nuc}}\bigg)^\frac{1}{3}=180m$
b) \frac{M}{\frac{4}{3}\piR^3}=\frac{m_U}{\frac{4}{3}\pir^3}
$r_U=R(\frac{m_U}{M})^\frac{1}{3}=2.58\times10^{-10}m$
$r_{actual}=(1.2\times10^{-15}m)(238)^\frac{1}{3}=7.44\times10^{-15}m$
$\frac{r_U}{r_{actual}}=3.47\times10^4$
