Explanation

(b)
The complex which contains 18 valence electrons, follows 18-electron rule.
(a) In $\lbrack V(CO)_{5}\rbrack$
The number of valence electrons
= 5+(2$\times$5)
= 15 $e^{-}$
(b) In$\left\lbrack Fe\left( {NH}_{3} \right)_{6} \right\rbrack^{2 +}$,
The number of valence electrons
=6+(6$\times$2)=6+12=18 $e^{-}$
(c) In $\lbrack Ni(CO)_{6}\rbrack$,
The number of valence electrons=10+(2$\times 6$)=22 $e^{-}$
(d) In $\left\lbrack Mn\left( H_{2}O \right)_{6} \right\rbrack^{2 +}$ ,
The number of valence electrons=5+(6$\times$2)=17 $e^{-}$
Thus, only $\left\lbrack Fe\left( {NH}_{3} \right)_{6} \right\rbrack^{2 +}$follows 18-electron rule.