Answer
$v = 0.866~c$
Work Step by Step
Let $t' = 1~s$ and let $t_0 = 0.5~s$. We can find the required speed in the moving reference frame:
$t' = \gamma~t_0$
$t' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}~t_0$
$\sqrt{1-\frac{v^2}{c^2}} = \frac{t_0}{t'}$
$1-\frac{v^2}{c^2} = (\frac{t_0}{t'})^2$
$\frac{v^2}{c^2} = 1-(\frac{t_0}{t'})^2$
$v^2 = [1-(\frac{t_0}{t'})^2] \times c^2$
$v = \sqrt{1-(\frac{t_0}{t'})^2}~\times c$
$v = \sqrt{1-(\frac{0.5~s}{1~s})^2}~\times c$
$v = \sqrt{0.75}~\times c$
$v = 0.866~c$
