Answer
(a) $v = 0.9965~c$
(b) $d = 119.6~ly$
Work Step by Step
(a) Let $t' = 120~y$ and let $t_0 = 10~y$. We can find the required speed in the moving reference frame:
$t' = \gamma~t_0$
$t' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}~t_0$
$\sqrt{1-\frac{v^2}{c^2}} = \frac{t_0}{t'}$
$1-\frac{v^2}{c^2} = (\frac{t_0}{t'})^2$
$\frac{v^2}{c^2} = 1-(\frac{t_0}{t'})^2$
$v^2 = [1-(\frac{t_0}{t'})^2] \times c^2$
$v = \sqrt{1-(\frac{t_0}{t'})^2}~\times c$
$v = \sqrt{1-(\frac{10~y}{120~y})^2}~\times c$
$v = 0.9965~c$
(b) We can find the distance to the distant star according to Mission Control:
$d = v~t$
$d = (0.9965~c)(120~y)$
$d = 119.6~ly$