In a communication system operating at wavelength 800 nm, only one percent of source frequency is available as signal bandwidth. The number of channels accomodated for transmitting TV signals of band width 6 MHz
are $($Take velocity of light $\mathrm{c} = 3 \times {10}^8\mathrm{m}/\mathrm{s},\mathrm{h} = 6.6\times {10}^{–34} \mathrm{J}-\mathrm{s})$

The communication system's supplied wavelength, according to the question, is 800 nm. Thus, using the following formula and the wavelength, we can determine the source frequency:

$$\begin{gathered} f=\frac{c}{\lambda } \\ \lambda =800\times {10}^{-9}m \\ f=\frac{3\times {10}^8}{800\times {10}^{-9}} \\ f=3.57\times {10}^{14}Hz \end{gathered}$$

Available frequency= $1\%$ of Source frequency

Available frequency= $3.75\times {10}^{14}Hz$

$1MHz={10}^{6}Hz$

Available frequency

$$\begin{gathered} =\frac{3.75\times {10}^{12}Hz}{{10}^6} \\ =3.73\times {10}^{6}MHz \end{gathered}$$

The formulas below provide the number of channels that can be used to transmit TV signals with a bandwidth of 6 MHz,

$$\begin{gathered} No. of channels=\frac{Available frequency}{Total bandwidth} \\ No. of channels=\frac{3.73\times {10}^6}{6} \\ No. of channels=6.25\times {10}^5 \end{gathered}$$

Hence the correct answer is $6.25\times {10}^5.$