An aluminium nucleus has 13 protons and 14 neutrons…

Step 1. The mass number. $A = Z + N = 13 + 14 = 27$.

Step 2. The radius. $R = 1.2\ \text{fm} \times 27^{1/3} = 1.2\ \text{fm} \times 3 = 3.6$ fm.

Step 3. The ratio. The other nucleus has $R = 1.2\ \text{fm} \times 216^{1/3} = 1.2\ \text{fm} \times 6 = 7.2$ fm. So the ratio is $\dfrac{3.6\ \text{fm}}{7.2\ \text{fm}} = 0.5$. We could also write $\left(\dfrac{27}{216}\right)^{1/3} = \dfrac{1}{2}$.