Answer
Relative to the station, cruiser B is required to travel at a speed of $~~0.500 c$
Work Step by Step
Let $u'$ be the velocity of cruiser A relative to the station
Then $u' = 0.800c$
Let $v$ be the velocity of the station relative to cruiser B
Then $v = -u$
We can find $u$, the velocity of cruiser A relative to cruiser B:
$u = \frac{u'+v}{1+u'~v/c^2}$
$u = \frac{0.800c-u}{1+(0.800c)(-u)/c^2}$
$u = \frac{0.800c^2-u~c}{c+(0.800)(-u)}$
$uc-0.800~u^2 = 0.800c^2-u~c$
$-0.800~u^2+2uc - 0.800c^2 = 0$
$0.800~u^2-2uc + 0.800c^2 = 0$
We can use the quadratic formula:
$u = \frac{2c\pm \sqrt{(-2c)^2-(4)(0.800)(0.800c^2)}}{(2)(0.800)}$
$u = \frac{2c\pm \sqrt{1.44c^2}}{1.60}$
$u = \frac{2c\pm 1.20c}{1.60}$
$u = 0.500c, 2c$
Since $2c$ is non-physical, the solution is $u = 0.500~c$
Therefore, relative to the station, cruiser B is required to travel at a speed of $~~0.500 c$
