Answer
$0.55c$
Work Step by Step
We can find the required speed as follows:
$v=\frac{c}{\sqrt{1+\frac{\Delta t_{\circ}^2c^2}{L_{\circ}^2}}}$
We plug in the known values to obtain:
$v=\frac{c}{\sqrt{1+\frac{(0.15\times 10^{-9}s)^2(3\times 10^8m/s)^2}{(3\times 10^{-2}m)^2}}}$
$v=0.55c$
