Step 1: Read what the data is telling us.
We are handed a Body-Centred Cubic (BCC) crystal of an element whose atomic mass is $M = 93$ and whose measured density is $\rho = 8.55\ \text{g cm}^{-3}$. Our target is the cube edge $a$ written as a cube root.
Step 2: Count the atoms in one BCC cell.
A BCC cell has 8 corner atoms shared by 8 cells plus 1 atom fully inside the body. So $z = 8 \times \frac{1}{8} + 1 = 2$ atoms per unit cell.
Step 3: Connect mass and volume.
One cell holds the mass of $z$ atoms, which is $\frac{z M}{N_A}$ grams, and occupies a volume $a^3$. Density is mass over volume, so $\rho = \frac{z M}{a^3 N_A}$.
Step 4: Make the edge the subject.
Rearranging gives $a^3 = \frac{z M}{\rho N_A}$.
Step 5: Put the numbers in.
$a^3 = \frac{2 \times 93}{8.55 \times 6.022 \times 10^{23}} = \frac{186}{51.49 \times 10^{23}}$.
Step 6: Simplify the arithmetic.
$\frac{186}{51.49} \approx 3.61$, so $a^3 \approx 3.61 \times 10^{-23}\ \text{cm}^3$.
Step 7: Take the cube root.
Therefore $a = (3.61 \times 10^{-23})^{1/3}\ \text{cm}$, which is option (1).
\[ \boxed{a = (3.61 \times 10^{-23})^{1/3}\ \text{cm}} \]