Pre lab calculations HELP!

[g1na]

right so this is relating to to my SAC using colorimetry to determine phosphorus in lawn fertiliser

in this experiment you will prepare a 250mL solution of fertiliser. It is desirable for the concentration of the phosphate ions in the sample to be about 50mg/L. if the fertiliser contains about 1.8% phosphorus by mass in form of water soluble phosphate, calculate the approximate mass of fertiliser required to prepare the 250mL solution

i'm not sure if it's right but i got an answer of 0.694g

[ggxoxo]

I got 6.944 g

Working:
0.05 g/L
0.0125 g in 250 mL

0.0018x=0.0125
x= 6.944444 g

[g1na]

 HAHAHAH yeah my bad, but for some bizarre reason, the teacher said that 0.23g was the answer but wont show us the working out of it :S