Step 1: Recall the Pauli commutation identity.
The three Pauli matrices obey $[\sigma_i, \sigma_j] = 2i\,\epsilon_{ijk}\,\sigma_k$, where $\epsilon_{ijk}$ is the Levi-Civita symbol (it is $+1$ for the cyclic order $x,y,z$, $-1$ for the reverse order, and $0$ if any index repeats). This is a standard algebraic fact about spin-$\tfrac12$ operators, and it saves us from multiplying matrices entry by entry.
Step 2: Apply the identity to $[\sigma_x, \sigma_y]$.
Here $i=x$, $j=y$, $k=z$, and $x,y,z$ is the cyclic order, so $\epsilon_{xyz} = 1$.
\[ [\sigma_x, \sigma_y] = 2i\,\sigma_z \]
Step 3: Multiply by $\sigma_z$.
\[ \sigma_z[\sigma_x,\sigma_y] = \sigma_z(2i\sigma_z) = 2i\,\sigma_z^2 \]
Every Pauli matrix squares to the identity, $\sigma_z^2 = I$, because each is both Hermitian and unitary. So $\sigma_z[\sigma_x,\sigma_y] = 2i\,I$.
Step 4: Take the trace.
The trace of the $2\times2$ identity matrix is $\text{Tr}(I) = 1+1 = 2$.
\[ \text{Tr}(\sigma_z[\sigma_x,\sigma_y]) = \text{Tr}(2i\,I) = 2i \times 2 = 4i \]
Step 5: Rule out the distractors.
$2i$ (option A) is the coefficient sitting in front of $\sigma_z$ in the commutator itself, before it gets multiplied by $\sigma_z$ again and traced, so picking it directly skips a step. $i$ and $i/2$ (options B and D) do not follow from this identity under any natural slip, so they are just there to catch factor-of-2 errors.
Final Answer:
Using the Pauli algebra identity, $\text{Tr}(\sigma_z[\sigma_x,\sigma_y]) = 4i$, option (C).\[ \boxed{4i} \]