Answer
At a speed of $~~3000~km/s$, the meter stick would shrink by a hair.
Work Step by Step
Let $L_0 = 1~m$
Let $L = 1-50~\mu m = 0.99995~m$
We can find the required speed $v$ of the meter stick:
$L = L_0~\sqrt{1-\frac{v^2}{c^2}}$
$\frac{L}{L_0} = \sqrt{1-\frac{v^2}{c^2}}$
$(\frac{L}{L_0})^2 = 1-\frac{v^2}{c^2}$
$\frac{v^2}{c^2} = 1- (\frac{L}{L_0})^2$
$v^2 = [1- (\frac{L}{L_0})^2]~c^2$
$v = \sqrt{1- (\frac{L}{L_0})^2}~c$
$v = \sqrt{1- (\frac{0.99995~m}{1~m})^2}~c$
$v = \sqrt{0.0000999975}~c$
$v = 0.00999987~c$
$v = (0.00999987)~(3.0\times 10^8~m/s)$
$v = 3000~km/s$
At a speed of $~~3000~km/s$, the meter stick would shrink by a hair.
