Magnetic attraction depends on number of unpaired electrons
The greater the number of unpaired electrons, the greater the paramagnetism and hence the greater the attraction towards an applied magnetic field.

We compare the number of unpaired electrons in each complex.

Option A: $\left[\mathrm{Ni}(\mathrm{H_2O})_6\right]^{2+}$

• Oxidation state of Ni = $+2$
• Electronic configuration of $\mathrm{Ni}^{2+}$: $\mathrm{Ni}: [Ar]3d^84s^2 \Rightarrow \mathrm{Ni}^{2+}: [Ar]3d^8$
• $\mathrm{H_2O}$ is a weak field ligand, so octahedral high-spin arrangement applies.
• For $d^8$ octahedral: $t_{2g}^6 e_g^2$
• Number of unpaired electrons = $2$

Option B: $\left[\mathrm{Co}(\mathrm{H_2O})_6\right]^{2+}$

• Oxidation state of Co = $+2$
• Electronic configuration of $\mathrm{Co}^{2+}$: $\mathrm{Co}: [Ar]3d^74s^2 \Rightarrow \mathrm{Co}^{2+}: [Ar]3d^7$
• $\mathrm{H_2O}$ is a weak field ligand, so this is a high-spin octahedral complex.
• For high-spin $d^7$ octahedral: $t_{2g}^5 e_g^2$
• Number of unpaired electrons = $3$

Option C: $\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$

• Oxidation state of Co = $+3$
• Electronic configuration of $\mathrm{Co}^{3+}$: $\mathrm{Co}^{3+}: 3d^6$
• $\mathrm{en}$ is a strong field ligand, especially with $\mathrm{Co}^{3+}$, so this is low-spin octahedral.
• For low-spin $d^6$ octahedral: $t_{2g}^6 e_g^0$
• Number of unpaired electrons = $0$

Option D: $\left[\mathrm{Zn}(\mathrm{H_2O})_6\right]^{2+}$

• Oxidation state of Zn = $+2$
• Electronic configuration of $\mathrm{Zn}^{2+}$: $\mathrm{Zn}: [Ar]3d^{10}4s^2 \Rightarrow \mathrm{Zn}^{2+}: [Ar]3d^{10}$
• Number of unpaired electrons = $0$

Comparison

• A: $2$ unpaired electrons
• B: $3$ unpaired electrons
• C: $0$ unpaired electrons
• D: $0$ unpaired electrons

Therefore, the complex with maximum attraction to magnetic field is: $\boxed{\left[\mathrm{Co}(\mathrm{H_2O})_6\right]^{2+}}$

Comparison with stored answer
Stored correct answer: B
Derived answer: B
They match.