Calculate balance between signal and noise to allow 32 kbit/s transfer
The signal-to-noise ratio needed for \$\mathrm{32\ kbit/s}\$ with a bandwith of \$\mathrm{100\ kHz}\$ is calculated in the following way:

\$\underbrace{C=B\cdot \log_2{\left(1+\dfrac SN\right)}}_{\textrm{Shannon's equation}}\longrightarrow \dfrac CB=\log_2{\left(1+\dfrac SN\right)}\longrightarrow 2^\frac{C}{B}=1+\frac SN\longrightarrow \frac SN=2^\frac CB - 1\$

If we substitute the values in the equation we get:

\$\dfrac SN=2^\frac{32\cdot 10^{3}\ \textrm{bits/s}}{10^5\ \textrm{Hz}} - 1=2^{0.32} - 1=0.2483\longrightarrow 10\cdot\log_{10}{(0.2483)}=-6.05\textrm{ dB}\$.



In Shannon's equation:


C is the channel capacity in bits per second;  
B is the bandwidth of the channel in hertz;  
S is the average signal power received over the bandwidth in watts;  
N is the average noise power received over the bandwidth in watts;
S/N is the signal-to-noise-ratio as a linear power ratio (not as decibels)




Sources:

https://en.wikipedia.org/wiki/Shannon%E2%80%93Hartley_theorem
https://www.st-andrews.ac.uk/~www_pa/Scots_Guide/iandm/part8/page1.html