Answer
See the answer below.
Work Step by Step
a) Cr (group 6, period 4):
$[Ar]\ 3d^5\ 4s^1$
Removing three electrons:
$Cr^{3+}$: $[Ar]\ 3d^3$
b) Both are paramagnetic since they have unpaired electrons, but $Cr^{2+}$ has 4 unpaired, thus being more paramagnetic.
c) Since radius increases from top to bottom on the periodic table, the radius of $Al^{3+}$ should be smaller than 64 pm since Al is in the third period.
