Answer
$2.60\times10^{8}\,m/s$
Work Step by Step
Recall that $L=L_{0}\sqrt {1-\frac{v^{2}}{c^{2}}}$ where $L_{0}$ is the original length, $L$ is the observed length, $v$ is the moving speed and $c$ is the speed of light.
Given $L=\frac{L_{0}}{2}$
$\implies \frac{1}{2}=\sqrt {1-\frac{v^{2}}{c^{2}}}$
$\implies \frac{1}{4}=1-\frac{v^{2}}{c^{2}}$
Or $\frac{v^{2}}{c^{2}}=1-\frac{1}{4}=\frac{3}{4}$
$\implies v=\frac{\sqrt 3}{2}c=\frac{\sqrt {3}}{2}\times(3.00\times10^{8}\,m/s)$
$=2.60\times10^{8}\,m/s$
