Answer
$v = 0.126~c$
$v = 3.78\times 10^7~m/s$
Work Step by Step
Let $L_0 = 1.0~m$ and let $L = 0.992~m$.
We can find the speed $v$ with respect to the lab:
$L = L_0~\sqrt{1-\frac{v^2}{c^2}}$
$\sqrt{1-\frac{v^2}{c^2}} = \frac{L}{L_0}$
$1-\frac{v^2}{c^2} = (\frac{L}{L_0})^2$
$\frac{v^2}{c^2} = 1-(\frac{L}{L_0})^2$
$v^2 = [1-(\frac{L}{L_0})^2]~(c^2)$
$v = \sqrt{1-(\frac{L}{L_0})^2}~\times c$
$v = \sqrt{1-(\frac{0.992~m}{1.0~m})^2}~\times c$
$v = 0.126~c$
$v = (0.126)~(3.0\times 10^8~m/s)$
$v = 3.78\times 10^7~m/s$
