One solved example in transmitter power?
Basically, this exercise tries to illustrate what is the meaning of the capacity channel by using the Shannon capacity channel as the simplest model.
The information we have is:

How fast we have transmitted a determined amount of information. 1 MB in 1 second
How noisy the channel is. 60 dB
How many losses the channel have. 174 dBm/Hz
How wide the channel is. 1 MHz

Now, we want to know the input power of the signal. With all data above it is possible to figure out.
First goal is compute output signal power.
Capacity channel is the maximum transmission data rate possible without error in the transmitted data. This parameter depends on the power signal, the amount of noise added by the channel and the bandwidth. \$ R = W \log_2{(1+ \frac{S}{N})}\$, where R is the capacity channel, W the bandwidth, S the power signal in linear units, N the floor noise in linear units. Assuming we are transmitting at the maximum rate the channel can handle, we can compute R as the amount of data transmitted (1 MB) divided by the time employed to transmit it:
$$R = \frac{1~MB }{1~s} \cdot \frac{10^6~byte}{1~MB} \cdot \frac{8~bit}{1~byte} = 8~Mb/s $$
According to noise, we know the spectral density noise (\$ N_0 \$), i.e. how much noise per Hz the channel is adding. To know the total noise power in the communication,  \$ N = N_0 \cdot B \$.
Now, we just need to compute \$ S_0 \$ in the capacity channel expression taking into account that logarithm is base 2.
$$R = W \log_2{(1+ \frac{S}{N})}$$
$$8\cdot 10^6 = 10^6 \log_2{\left(1+\frac{S_0}{4\cdot 10^{-15}}\right)}$$
$$\frac{8\cdot 10^6}{10^6} = \log_2{\left(1+\frac{S_0}{4\cdot 10^{-15}}\right)} $$
$$8 = \log_2{\left(1+\frac{S_0}{4\cdot 10^{-15}}\right)} $$
Now, using the logarithm property \$ a = \log_b{x} \Rightarrow b^a = x\$
$$2^8 = 1+\frac{S_0}{4\cdot 10^{-15}} $$
$$256 = 1+\frac{S_0}{4\cdot 10^{-15}} $$
$$255 = \frac{S_0}{4\cdot 10^{-15}} $$
$$255 \cdot 4\cdot 10^{-15} = S_0 $$
$$1.02\cdot 10^{-12} = S_0$$
As \$S_0\$ is power, the conversion to dBm is:
$$S_0~(dBm) = 10\log_{10}{\frac{1.02\cdot 10^{-12}}{10^{-3}}} \approx -90~ dBm$$
Finally, as the output signal is 60 dB smaller at the output than at the input due to channel losses:
$$S_0 = S_i - L$$
$$S_i = S_0 + L = -90 - \left( -60\right) = -90 + 60 = -30~dBm = 10^{\frac{-30-30}{10}} = 10^{-6}~W = 1~\mu W$$
(Remember that dBm to W is \$ 10^{\frac{Power_{dBm}-30}{10}} \$)