$(1)$ $C r^{3+} \rightarrow 3 d^1 4 s^2$ no. of unpaired $e^{-}=1$

$(2)$ $M n^{3+} \rightarrow 3 d^3 4 s^2$ no. of unpaired $e^{-}=1$

$(3)$ $F e^{3+} \rightarrow 3 d^3 4 s^2 \quad$ no. of unpaired $e^{-}=1$

$(4)$ $Co ^{3+} \rightarrow 3 d^6 4 s^2$ no. of unpaired $e^{-}=0$

So $C o$ in $\left[\operatorname{Co}(C N)_6\right]^{3-}$ is having less number of unpaired electrons

comparatively i.e 0 . So it is a diamagnetic compound means less paramagnetic comparatively.