Explanation

(d)
In ${MnSO}_{4}\ .4H_{2}O$, $Mn$ is present as ${Mn}^{2 +}$
${Mn}^{2 +}$= $3d^{5}$ $4s^{0}$
[table] $\upharpoonleft$ | $\upharpoonleft$ | $\upharpoonleft$ | $\upharpoonleft$ | $\upharpoonleft \ \ \$ | | [/table]
(Unpaired electrons =5)
In ${CuSO}_{4}.5H_{2}O$, Cu is present as ${Cu}^{2 +}$
${Cu}^{2 +}$= $3d^{9}$ $4s^{0}$
[table] ⥮ | ⥮ | ⥮ | ⥮ | $\upharpoonleft$ | | [/table]
(Unpaired electrons =1)
In ${FeSO}_{4}.6H_{2}O$,Fe is present as ${Fe}^{2 +}$
${Fe}^{2 +} = \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3d^{6}$ $4s^{0}$
[table] ⥮ | $\upharpoonleft$ | $\upharpoonleft$ | $\upharpoonleft$ | $\upharpoonleft$ | | [/table]
(Unpaired electrons =4)
In ${NiSO}_{4}.6H_{2}O$ Ni is present as ${Ni}^{2 +}$
${Ni}^{2 +}$= $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3d^{8}$ $4s^{0}$
[table] ⥮ | ⥮ | ⥮ | $\upharpoonleft$ | $\upharpoonleft$ | | [/table]
(Unpaired electrons =2)
Since, paramagnetic character $\propto \$unpaired electrons .
Thus, ${CuSO}_{4}.5H_{2}O$ has the lowest degree of paramagnetism among the given at 298 K.