$54.64$ $"g"$ anhydrate
With
Mols are interconvertible, as one mol of anything is the same number of particles as one mol of anything else.
$100.0 cancel("g FeSO"_4cdot7"H"_2"O") xx ("1 mol FeSO"_4cdot7"H"_2"O")/(278.02 cancel("g FeSO"_4cdot7"H"_2"O"))$
$=$ $"0.3597 mols"$
For each
$0.3597 cancel("mols FeSO"_4cdot7"H"_2"O") xx "7 mols water"/cancel("1 mol FeSO"_4cdot7"H"_2"O")$
$=$ $"2.518 mols water"$
Consequently, you have this mass of water by using its molar mass:
$2.518 cancel"mols water" xx ("18.015 g water")/cancel"mol water"$
$=$ $"45.36 g water"$
So, your final mass should be