The correct order of the wavelength maxima of the absorption band in the ultraviolet-visible region for the given complexes is

For octahedral $$d^6$$ cobalt(III) complexes the chief electronic transition seen in the UV-visible spectrum is the spin-allowed $$t_{2g}^6e_g^0 \rightarrow t_{2g}^5e_g^1$$ (a $$d-d$$ transition).
The energy of this transition equals the crystal-field splitting energy $$\Delta_0$$; hence

$$E = h\nu = \frac{hc}{\lambda_{\max}} = \Delta_0 \; \Longrightarrow \; \lambda_{\max} \propto \frac{1}{\Delta_0}$$

So, the larger the splitting $$\Delta_0$$, the smaller (bluer) is the absorption wavelength, and vice-versa.

Because the metal ion is the same (Co(III)) and the geometry is the same (octahedral), the magnitude of $$\Delta_0$$ depends almost entirely on the ligand field strength. The relevant order from the spectrochemical series is

$$CN^- \gt NH_3 \gt H_2O \gt Cl^-$$

Therefore, the descending order of $$\Delta_0$$ (strongest field to weakest field) for the given complexes is:

$$[Co(CN)_6]^{3-} \; \gt \; [Co(NH_3)_6]^{3+} \; \gt \; [Co(NH_3)_5(H_2O)]^{3+} \; \gt \; [Co(NH_3)_5(Cl)]^{2+}$$

Converting this into the ascending order of the absorption wavelength $$\lambda_{\max}$$ (since $$\lambda_{\max} \propto 1/\Delta_0$$) gives:

$$[Co(CN)_6]^{3-} \; \lt \; [Co(NH_3)_6]^{3+} \; \lt \; [Co(NH_3)_5(H_2O)]^{3+} \; \lt \; [Co(NH_3)_5(Cl)]^{2+}$$

Thus the correct sequence matches Option A.

Option A which is: [Co(CN)$$_6$$]$$^{3-}$$ < [Co(NH$$_3$$)$$_6$$]$$^{3+}$$ < [Co(NH$$_3$$)$$_5$$(H$$_2$$O)]$$^{3+}$$ < [Co(NH$$_3$$)$$_5$$(Cl)]$$^{2+}$$