Okay, 1.9 + 1.9 moles were put in, making 3.8 moles altogether.
So if 3.0 moles ended up as HI then the other 0.8 must be H2 and I2 (so 0.4 of each)
Converting this to concentration in mol dm-3 by dividing by the volume (0.25 dm3) gives 12 for [HI], 1.6 for and 1.6 for [I2].
Therefore Kc = ([HI]^2) / ([I2]) = (12^2) / (1.6*1.6) = 144/2.56 = 56.25 arbitrary units
I know that's not the answer you gave but I think that answer is wrong, the units definitely are because the Kc should be unitless in this case