Answer
$v=0.76c$
Work Step by Step
We can find the required speed as follows:
$L=L_{\circ}\sqrt{1-v^2/c^2}$
This simplifies to:
$v=c\sqrt{1-L^2/L_{\circ}^2}$
We plug in the known values to obtain:
$v=3.00\times 10^8\sqrt{1-(80.5/124)^2}$
$v=0.76c$
