Answer
The thickness of the strip is $8.1\times 10^{-7}~m$
Work Step by Step
We can write an expression for the drift speed:
$v_d = \frac{I}{n~q~A} = \frac{I}{n~q~W~T}$
$I$ is the current
$n$ is the number of electrons per unit of volume
$q$ is the charge of one electron
$A$ is the cross-sectional area
$W$ is the cross-sectional width
$T$ is the cross-sectional thickness
We can find the thickness $T$:
$v_d = \frac{I}{n~q~W~T}$
$T = \frac{I}{n~q~W~v_d}$
$T = \frac{130 \times 10^{-6}~A}{(8.8\times 10^{22}~m^{-3})(1.6\times 10^{-19}~C)~(260\times 10^{-6}~m)(44~m/s)}$
$T = 8.1\times 10^{-7}~m$
The thickness of the strip is $8.1\times 10^{-7}~m$
