Answer
$8.4\times 10^{-16}$
Work Step by Step
We know that
$\frac{V_{nucleus}}{V_{atom}}=\frac{(4/3)\pi r_n^3}{(4/3)\pi r_a^3}$
$\implies \frac{V_{nucleus}}{V_{atom}}=(\frac{r_n}{r_a})^3$
We plug in the known values to obtain:
$\frac{V_{nucleus}}{V_{atom}}=(\frac{0.50\times 10^{-15}}{5.3\times 10^{-11}})^3=8.4\times 10^{-16}$