Step 1: Identify the given radii and determine the crystal structure. The radius of the cation A$^+$ ($r_c$) is $1.0 \, \text{pm}$. The radius of the anion X$^-$ ($r_a$) is $2.0 \, \text{pm}$. Calculate the radius ratio $\frac{r_c}{r_a}$: $$\frac{r_c}{r_a} = \frac{1.0 \, \text{pm}}{2.0 \, \text{pm}} = 0.5$$ For ionic compounds with a radius ratio in the range $0.414 \le \frac{r_c}{r_a} < 0.732$, the coordination number is 6. This typically corresponds to an NaCl-type crystal structure, where the anions form a Face-Centered Cubic (FCC) lattice and the cations occupy the octahedral voids. Step 2: Determine the relationship between the unit cell edge length and the ionic radii for an NaCl-type structure. In an NaCl-type structure, the cation and anion touch along the edge of the unit cell. The edge length ($a$) is twice the sum of the cation and anion radii. $$a = 2(r_c + r_a)$$ Step 3: Calculate the edge length ($a$) of the unit cell. $$a = 2(1.0 \, \text{pm} + 2.0 \, \text{pm})$$ $$a = 2(3.0 \, \text{pm})$$ $$a = 6.0 \, \text{pm}$$ Step 4: Calculate the volume of the unit cell ($V$). The volume of a cubic unit cell is given by $V = a^3$. $$V = (6.0 \, \text{pm})^3$$ $$V = 216 \, \text{pm}^3$$ Step 5: Compare the calculated volume with the given options. The calculated volume is $216 \, \text{pm}^3$, which matches option (d). The final answer is $\boxed{\text{d) } 216 \, \text{pm}^3}$.