I am not going to use the numbers you gave me because it is my preference to do a problem in general; you can stick the numbers in.  The speed of the space ship as seen by the space station I will take as , the speed of the laser in both frames is c, the speed of the rocket in the space ship frame I will take as   (Note that I assume that the space ship is moving toward the space station; if it were moving away you would have to replace β by β everywhere.)  The speed of the rocket in the space station frame is obtained from the velocity addition formula:

                        .

In the space station frame, the distance which must be traveled I will take to be L (the distance to the ship when it fires.  So it is now straightforward to calculate the times t (the time for the laser) and tR (the time for the rocket):

                         

and

                        .

Next we find the times  and  as seen by the space ship.  Two things have to be taken into consideration: the distance between the two is length-contracted, , and in the time  (or  ) the space station moves up a distance  (or  ) to “meet” the laser (or rocket).  So, now,

                        ;

and

                        .