Using the formula:

$ I = n e A v $

where $I$ is the current, $n$ is the number density of free electrons, $e$ is the charge of an electron, $A$ is the cross-sectional area of the wire, and $v$ is the drift speed of the electrons.

We can isolate $v$ to find:

$ v = \frac{I}{n e A}$

Substituting the given values:

$ v = \frac{3.2 \, \text{A}}{8 \times 10^{28} \, \text{m}^{-3} \times 1.6 \times 10^{-19} \, \text{C} \times 2 \times 10^{-6} \, \text{m}^2}$

This simplifies to:

$ v = \frac{3.2}{16 \times 1.6 \times 10^3} \, \text{ms}^{-1} = 125 \times 10^{-6} \, \text{ms}^{-1} $

So, the drift speed of the electrons is indeed $125 \times 10^{-6} \, \text{ms}^{-1}$.