The correct increasing order of spin-only magnetic moment values of...

Calculate spin-only magnetic moment using $\mu_s = \sqrt{n(n+2)}$ where $n$ is the number of unpaired electrons.
For $[\text{MnBr}_4]^{2-}$ (A): Mn(II) in tetrahedral environment is high spin with 5 unpaired electrons. $\mu_s = \sqrt{5(7)} = \sqrt{35} \approx 5.92$ BM.
For $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ (B): Cu(II) has 1 unpaired electron. $\mu_s = \sqrt{1(3)} = \sqrt{3} \approx 1.73$ BM.
For $[\text{Ni}(\text{CN})_4]^{2-}$ (C): Ni(II) in square planar geometry is low spin with 0 unpaired electrons. $\mu_s = 0$ BM.
For $[\text{Ni}(\text{H}_2\text{O})_6]^{2+}$ (D): Ni(II) in octahedral environment is high spin with 2 unpaired electrons. $\mu_s = \sqrt{2(4)} = \sqrt{8} \approx 2.83$ BM.
Increasing order: C (0) < B (1.73) < D (2.83) < A (5.92)