Step 1: Recall the density formula for a crystalline unit cell.
\[ \rho = \frac{Z \times M}{a^3 \times N_A} \] where $Z$ = atoms per unit cell, $M$ = molar mass, $a$ = edge length, $N_A$ = Avogadro number.
Step 2: Identify given values and $Z$ for fcc lattice.
For fcc: $Z = 4$. Given: $\rho = 8.92$ g/cm$^3$, $a = 3.61 \times 10^{-8}$ cm, $N_A = 6.022 \times 10^{23}$ mol$^{-1}$.
Step 3: Rearrange to solve for molar mass $M$.
\[ M = \frac{\rho \times a^3 \times N_A}{Z} \]
Step 4: Calculate $a^3$.
\[ a^3 = (3.61)^3 \times 10^{-24}\ \text{cm}^3 \] $(3.61)^3 = 47.05$. So $a^3 = 47.05 \times 10^{-24}$ cm$^3$.
Step 5: Substitute and calculate $M$.
\[ M = \frac{8.92 \times 47.05 \times 0.6022}{4} = \frac{252.7}{4} \approx 63.18\ \text{g mol}^{-1} \] This corresponds to copper (Cu), consistent with the given density and lattice parameter.
Step 6: State the final answer.
\[ \boxed{63.178\ u} \]