Answer
The Earth's diameter contracts by $~~63.7~mm$
Work Step by Step
We can find $\gamma$:
$\gamma = \frac{1}{\sqrt{1-(v/c)^2}}$
$\gamma = \frac{1}{\sqrt{1-(\frac{3.0\times 10^4~m/s}{3.0\times 10^8~m/s})^2}}$
$\gamma = 1.000000005$
We can find the diameter according to an observer:
$d = \frac{d_0}{\gamma}$
$d = \frac{12,740~km}{1.000000005}$
$d = 12,739.9999363~km$
We can find the amount that the diameter is contracted:
$d_0-d = 0.0000637~km$
$d_0-d = 0.0637~m$
$d_0-d = 63.7~mm$
The Earth's diameter contracts by $~~63.7~mm$
