Use g = 14.0 for Fe³⁺ and the ligand factors f: fluoride 0.90, water 1.00 and cyanide 1.70…

Step 1. Fluoride. $\Delta_o = 0.90 \times 14.0 \times 1000 = 12600\ \text{cm}^{-1}$, so $\lambda = 10^7/12600 = 794\ \text{nm}$.

Step 2. Water. $\Delta_o = 14000\ \text{cm}^{-1}$, so $\lambda = 714\ \text{nm}$.

Step 3. Cyanide. $\Delta_o = 1.70 \times 14.0 \times 1000 = 23800\ \text{cm}^{-1}$, so $\lambda = 420\ \text{nm}$.