Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
To find the average density of an aluminum atom, we need to calculate the mass of the atom and then divide it by its volume.
$$\rho=\dfrac{M}{V}=\dfrac{27\;\rm u}{\frac{4}{3}\pi r^3}$$
Plug the known;
$$\rho=\dfrac{27\times 1.67\times 10^{-27}}{\frac{4}{3}\pi \left(\frac{1.2}{2}\times 10^{-10}\right)^3}$$
$$\rho=\color{red}{\bf4 .98\times 10^4}\;\rm kg/m^3$$
$$\color{blue}{\bf [b]}$$
The average volume per atom is given by the density
$$V_{\rm avg}=\dfrac{M}{\rho_{\rm Al}}=\dfrac{27\;\rm u}{2700}=\dfrac{27\times 1.67\times 10^{-27}}{2700}$$
$$V_{\rm avg}=\color{red}{\bf 1.67\times 10^{-29}}\;\rm m^3$$
So, the radius of the sphere here is then given by
$$\frac{4}{3}\pi r^3=V_{\rm avg} $$
So,
$$r=\sqrt[3]{\dfrac{V_{\rm avg} }{\frac{4}{3}\pi}}$$
$$r=\sqrt[3]{\dfrac{ 1.67\times 10^{-29} }{\frac{4}{3}\pi}}$$
$$r=\color{red}{\bf 1.6 \times 10^{-10}}\;\rm m $$
$$\color{blue}{\bf [c]}$$
To find the density of the aluminum nucleus, we'll calculate it using the mass and volume of the nucleus.
$$\rho_{\rm nucleus}=\dfrac{M}{V_{\rm nucleus}}=\dfrac{27\;\rm u}{\frac{4}{3}\pi r_{\rm nucleus }^3}$$
Plug the known;
$$\rho_{\rm nucleus}=\dfrac{27\times 1.67\times 10^{-27}}{\frac{4}{3}\pi \left(\frac{8}{2}\times 10^{-15}\right)^3}$$
$$\rho_{\rm nucleus}=\color{red}{\bf 1.682\times 10^{17}}\;\rm kg/m^3$$
To find how many times this density of the nucleus is greater than the solid aluminum;
$$\dfrac{\rho_{\rm nucleus}}{\rho}=\dfrac{1.682\times 10^{17}}{2700}=\bf 6.23\times 10^{13}$$
Hence,
$$\rho_{\rm nucleus}=\color{red}{\bf 6.23\times 10^{13}}\;\rho$$
