A. Sampling rates are the number of samples per second in a sound.

Here, Sample/Hz = $2$

Frequency = $100 \text{ kHz}$

∴ $\color{maroon}{\text{Minimum sampling rate}}$ = $2 \times 100,000$ samples/sec

$\qquad \qquad \qquad \qquad = \color{purple}{200,000 \text{ samples/sec }}$

B. Now, we'll determine the maximum data rate or channel capacity

Channel capacity: The maximum rate at which data can be transmitted over a given channel. Usually measured in bps (bits per sec)

∴ $\color{lightblue}{\text{Channel Capacity}}$ = $200,000 \text{ samples/sec} \times 16 \text{ bit/sample}$

$\qquad \qquad \qquad = 32,00,000 \text{ bps}$

$\qquad \qquad \qquad = \color{lightgreen}{3.2 \text{ Mbps}}$ $\left [ \text{1 Mbps = 1000000 bit Per second} \right]$

C. According to Shannon's Theorem

$C = B * log_2 (1 + \dfrac{S}{N})$

Where $B$= Bandwidth

$\dfrac{S}{N}$ = signal to noise ratio, expressed in dB

$C$ = maximum channel capacity

∴ $\dfrac{C}{B} = log_2(1 + \dfrac{S}{N})$

Or, $1+\dfrac{S}{N} = 2^{C/B} $

Or, $\dfrac{S}{N} = 2^{C/B} -1 $

Here, Channel Capacity or C will be $3.2 \text{ Mbps}$

Bandwidth or B will be $500 kHz$

∴ $\dfrac{S}{N} = 2^{C/B} -1 $

Or, $\dfrac{S}{N} = 2^{3200/500} -1$ $\left [ \text{3.2 Mbps = 3200 Kilobit per sec } \right ]$

Or, $\dfrac{S}{N} = 2^{6.4} -1$

Or, $\dfrac{S}{N} = 84.4485063 -1$

Or, $\dfrac{S}{N} = 83.4485063$

Now, $\dfrac{S}{N} \text{ in dB} = 10 \times log_{10}(\dfrac{S}{N})$

Or, $\dfrac{S}{N} \text{ in dB} = 10 \times log_{10} \left ( 83.4485063 \right)$

Or, $\dfrac{S}{N} \text{ in dB} = 10 \times (1.9213743)$

Or, $\dfrac{S}{N} \text{ in dB} = 19.213743 \text{ dB}$

∴ $\color{green}{\text{Minimum signal to noise ratio (in dB) will be}} \color{Gold}{\text{19.21 dB}}$ $\color{green}{\text{ is required to sustain the given data rate}} \color{green}{\text{ over a 500kHz radio channel. }}$