If \$N\$ is the number of bits then your input signal range is divided into quantization intervals of size
$$q = \frac{V_{ref}}{2^N}$$
The maximum quantization error is \$q/2\$ and it is usually assumed that the quantization error is uniformly distributed between \$-q/2\$ and \$q/2\$. So the PDF of the quantization error is constant between \$-q/2\$ and \$q/2\$ with height \$1/q\$. With this assumption the quantization noise power is
$$P_q = \frac{1}{q}\int_{-q/2}^{q/2}x^2dx = \frac{q^2}{12} = \frac{V_{ref}^2}{12\cdot 2^{2N}}$$
The signal power is given by
$$P_x = \frac{V_{max}^2}{2} \text{ with } V_{max} = 100mV$$
The factor \$1/2\$ in the signal power comes from the fact that the input is sinusoidal, i.e. its power is given by half its maximum value. Note that the maximum value is the amplitude which is half the peak-to-peak value. Putting everything together we get
$$SNR = 10\log\frac{P_x}{P_q} = 10\log\frac{6V_{max}^22^{2N}}{V_{ref}^2} = \\
= 10\log\frac{6V_{max}^2}{V_{ref}^2} + 10\log 2^{2N} =
10\log\frac{6(0.1)^2}{5^2} + N\cdot 20\log 2 =\\
= -26.2 + 12\cdot 6.02 = 46\text{dB}$$
EDIT: To see how the \$1.76\$dB pop up we now use \$V_{pp}\$ instead of \$V_{max}\$, where \$V_{pp}=2V_{max}\$ is the peak-to-peak input voltage. The SNR can then be written as
$$SNR = 10\log\frac{3V_{pp}^2}{2V_{ref}^2} + 10\log 2^{2N} =\\
= 10\log\frac{V_{pp}^2}{V_{ref}^2} + 10\log\frac{3}{2} + N\cdot 20\log 2 =\\
= 10\log\frac{V_{pp}^2}{V_{ref}^2} + 1.76 + 6.02\cdot N$$
So if we use the maximum input range, i.e. \$V_{pp}=V_{ref}\$ we get the formula that was mentioned in the question.
Spehro's comment is basically the answer. Imagine a link which subtracts a random, varying noise value between 0 and 0.5V from the input. You measure 4.5V at one end, but that could mean either a 4.5V input with no noise or 5V minus noise. You can't tell, and have lost your signal entirely.
With a digital signal varying between 5V and 0V over the same link, you can declare everything over 2.5V to be a '1', and lose no signal.
Best Answer
SNR as you know means Signal-to-Noise ratio. So it's \$20 log(\$\$V_S\over V_N\$), where \$V_S\$ is signal voltage and \$V_N\$ is noise voltage. So, if noise (\$V_N\$) is really small (compared to signal), then SNR gets really big.
It's quite possible to have systems with negative SNR (where the signal level is less than the noise), where you need synchronous demodulation or a lock-in amplifier to pull the signal out of the noise. In that case, you could write Noise-to-Signal ratio \$20 log(\$\$V_N\over V_S\$), which would be equal to - \$20 log(\$\$V_S\over V_N\$).
GNSS systems such as GPS have a negative SNR before processing.