Answer
$\rho = 2.3 \times 10^{17} kg/m^3$
Work Step by Step
The radius of nucleus is given by
$r = r_0 A^{1/3} $
To find the density, $\rho$
$\rho = \frac{M}{V}$
$\rho = \frac{M}{\frac{4}{3} \pi r^3 }$
substitute $r = r_0 A^{1/3} $ and $M = Au $ into the $r$ in density equation
$\rho = \frac{Au}{\frac{4}{3} \pi (r_0 A^{1/3})^3 }$
simplify the equation
$\rho = \frac{1u}{\frac{4}{3} \pi (r_0^3) }$
Now solve for $\rho$
$\rho = \frac{1.66\times 10^{-27} kg}{\frac{4}{3} \pi (1.2 \times 10^{-15} m)^3 }$
$\rho = 2.3 \times 10^{17} kg/m^3$
