C

Step 1: Drift velocity can be defined as the average velocity that a particle, such as an electron, attains due to an electric field.
Step 2: The drift velocity ($v_d$) is given by the formula:
$$ v_d = \frac{I}{nqA} $$
where:
- $I$ is the current,
- $n$ is the number of charge carriers per unit volume,
- $q$ is the charge of the electron (approximately $1.6 \times 10^{-19}$ C),
- $A$ is the cross-sectional area of the conductor.
Step 3: In metals, the number of free electrons ($n$) can be roughly around $10^{28} m^{-3}$.
Step 4: By substituting realistic values (for example, $I \approx 1A$, $A \approx 1 \times 10^{-6} m^2$), we can find that $v_d \approx 10^{-4} m/s$.
Step 5: The order of magnitude, thus, is approximately $10^{-4} m/s$.
So the order of magnitude of drift velocity of free electrons is approximately $10^{-4}$ m/s.
Therefore, the answer is C.