Only 2 % of the optical source frequency is the available channel...

Frequency $f = \dfrac{c}{\lambda} = \dfrac{3 \times {10}^{8}}{1000 \times {10}^{- 9}} = 3 \times {10}^{14}Hz$

Band width used for transmission $= \dfrac{2}{100} \times 3 \times {10}^{14}Hz = 6 \times {10}^{12}Hz$

Number of channels $N = \dfrac{\text{Total bandwidth of channel}}{\text{Bandwidth needed per channel}} = \dfrac{6 \times {10}^{12}}{8 \times {10}^{3}} = \dfrac{600}{8} \times {10}^{7} = 75 \times {10}^{7}$