Spin only magnetic moment is given by
\(\begin{array}{l}
\mu = \sqrt{n(n+2)} \; B.M.
\end{array}\)
where \(n\) = number of unpaired electrons.
\(Co^{3+}\)
\(\begin{array}{l}
Co : [Ar]\,3d^7 4s^2
\end{array}\)
\(\begin{array}{l}
Co^{3+} : 3d^6
\end{array}\)
\(n = 4\)
\(\begin{array}{l}
\mu = \sqrt{4(4+2)} = \sqrt{24}\;B.M.
\end{array}\)
\(Cr^{3+}\)
\(\begin{array}{l}
Cr : [Ar]\,3d^5 4s^1
\end{array}\)
\(\begin{array}{l}
Cr^{3+} : 3d^3
\end{array}\)
\(n = 3\)
\(\begin{array}{l}
\mu = \sqrt{3(3+2)} = \sqrt{15}\;B.M.
\end{array}\)
\(Fe^{3+}\)
\(\begin{array}{l}
Fe : [Ar]\,3d^6 4s^2
\end{array}\)
\(\begin{array}{l}
Fe^{3+} : 3d^5
\end{array}\)
\(n = 5\)
\(\begin{array}{l}
\mu = \sqrt{5(5+2)} = \sqrt{35}\;B.M.
\end{array}\)
\(Ni^{2+}\)
\(\begin{array}{l}
Ni : [Ar]\,3d^8 4s^2
\end{array}\)
\(\begin{array}{l}
Ni^{2+} : 3d^8
\end{array}\)
\(n = 2\)
\(\begin{array}{l}
\mu = \sqrt{2(2+2)} = \sqrt{8}\;B.M.
\end{array}\)
Hence, correct matching is
\(\begin{array}{l}
a \rightarrow iv
\\
b \rightarrow v
\\
c \rightarrow ii
\\
d \rightarrow i
\end{array}\)