The correct option is A [Cr(CN)6]3−
As we know,
magnetic moment
μ=√n(n+2) BM
n = no of unpaired electron so
higher no of unpaired electron will have higher moment.
[Co(CN)6]3−
+3 n = 4
[Fe(CN)6]3−
+3 n = 5
[Mn(CN)6]3−
+3 n = 4
[Cr(CN)6]3−
+3 n = 3
so it have lowest magnetic moment and shows least paramagnetic behaviour.