The correct order of ligands in the spectrochemical series is

The spectrochemical series ranks ligands from weak-field (produce the smallest crystal-field splitting $$\Delta$$) to strong-field (produce the largest $$\Delta$$).
A concise form of the series up to the ligands needed here is

$$I^- \lt Br^- \lt S^{2-} \lt SCN^- (S\;{\text{donor}}) \lt Cl^- \lt F^- \lt H_2O \lt NCS^- (N\;{\text{donor}}) \lt NH_3 \lt en \lt NO_2^- \lt CN^- \lt CO$$

Now locate the four ligands mentioned in the question:

• $$CN^-$$ is one of the very strongest field ligands because it is a good $$\sigma$$-donor and an excellent $$\pi$$-acceptor.
• en (ethylenediamine) is a neutral bidentate ligand. It donates two $$\sigma$$ pairs of electrons and additionally benefits from the chelate effect, so it is stronger than N-bound $$NCS^-$$ but weaker than $$CN^-$$.
• $$NCS^-$$ bound through nitrogen (as is usual in octahedral complexes) is weaker than en because it is a poorer $$\pi$$-acceptor and lacks the chelation advantage.
• $$Cl^-$$ is a large, weak $$\sigma$$-donor with no $$\pi$$ interactions, so it lies far toward the weak-field end.

Therefore the correct increasing-to-decreasing order of crystal-field strength (strongest first) is

$$CN^- \; \gt \; \text{en} \; \gt \; NCS^- \; \gt \; Cl^-$$

Comparing with the options, this matches

Option B which is: $$CN^- \, \gt \, \text{en} \, \gt \, NCS^- \, \gt \, Cl^-$$