Answer
$\beta = 0.866$
Work Step by Step
Given that the length of a spaceship is measured to be exactly half its rest length,
$ L =\frac{L_{o}}{2} $
The speed parameter $\beta$ is obtained from the length;
$L =L_{o} \sqrt (1-\beta^2)$
$\frac{L_{o}}{2} = L_{o}\sqrt (1-\beta^2)$
$\frac{1}{4} = (1-\beta^2)$
$\beta^2 =1-\frac{1}{4}$
$\beta = 0.866$
