I would leave it in km/hr since the plane's speed and wind are in the same units.
(one thing to be careful with here is the wind direction.....is it BLOWING in the direction of 250 degrees or is it COMING FROM 250 degrees.....as a pilot the latter would be more likely correct, but the way the question is worded it sounds like the former....the resultant ground speed will be the same BUT the resultant angle of the vector will be DIFFERENT) I will proceed with the wind BLOWING in the direction of 250 degrees:
Break both the plane and wind speeds into vertical and horizontal components
and sum them up.
Planex = 500 cos 135 = - 353.5533
Planey = 500 sine135 = 353.5533
Windx = 120 cos 250 = - 41.042
Windy = 120 sine250 = -112.764
Resultant x = -353.5533 - 41.024 = -394.577
Resultant y = 353.5533 - 112.764 = 240.789
Use pythagorean theorem x^2 + y^2 = resultant^2 Resultant = 462.245 KM/HR
With the x and y resultants you can figure out the new angle too (if the pilot does not correct by turning into the wind and 'crabbing' a bit) .....do you know how? (HINT: Tangent)
~jc
(hope I did all of the calculator math correctly!)