Calculate the radius ratio (rc)/(ra):

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.

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.

$$a = 2(r_c + r_a)$$

$$a = 2(1.0 \, \text{pm} + 2.0 \, \text{pm})$$ $$a = 2(3.0 \, \text{pm})$$ $$a = 6.0 \, \text{pm}$$

The volume of a cubic unit cell is given by $V = a^3$.

$$V = (6.0 \, \text{pm})^3$$ $$V = 216 \, \text{pm}^3$$

The calculated volume is $216 \, \text{pm}^3$, which matches option (d).

The final answer is $\text{d) } 216 \, \text{pm}^3$.