Answer
Electron drift speed is $5\times10^{-5}ms^{-1}$
Work Step by Step
Recall the equation,
$$I=nAVq$$
Where,
$n$ - Electron density of the material
$A$ - Surface area of the conductor
$V$ - Electron drift speed
$q$ - Charge of the electron
Here since the two wires are connected in series the current is the same,
$$nA_{1}V_{1}q=nA_{2}V_{2}q$$
$π(\frac{0.001m}{2})^{2}\times2.0\times10^{-4}ms^{-1}=π(\frac{0.002m}{2})^{2}\times V$
$$V=5\times10^{-5}ms^{-1}$$
