Answer
According to passengers on the spaceship, the passage takes $~~11.2~hours$
Work Step by Step
We can find the speed of the ship:
$v = \frac{d}{t}$
$v = \frac{10~light~hours}{15~hours}$
$v = \frac{2}{3}~c$
We can find $\gamma$:
$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$
$\gamma = \frac{1}{\sqrt{1-\frac{(2c/3)^2}{c^2}}}$
$\gamma = \frac{1}{\sqrt{1-\frac{4}{9}}}$
$\gamma = 1.34$
Let $t'$ be the time measured on earth.
We can find the time that passes according to passengers on the spaceship:
$t' = \gamma~t$
$t = \frac{t'}{\gamma}$
$t = \frac{15~h}{1.34}$
$t = 11.2~hours$
According to passengers on the spaceship, the passage takes $~~11.2~hours$
