Calculating signal/noise ratio and the noise level
The decibels (dB) provides a logarithmic scale for the ratio of two power values. Mathematically it is defined as:
$$
  R = 10 \log_{10} \left(\frac{P}{P_0}\right)
$$
where $P$ and $P_0$ are power values in linear scale (such as Watts) and $R$ is the ratio expressed in logarithmic scale. Often a fixed reference power level (such as 1 Watt) is chosen for $P_0$. When that reference power level $P_0$ is 1 milliwatt (mW), the resulting unit for the $R$ quantity is called a dBm.

Given a known reference level, you can conversely obtain the power in linear scale from the power ratio value expressed in logarithmic scale by using the inverse formula:
$$
  P = 10^{R/10}\cdot  P_0 
$$

So for a measurement of -46dBm (remember that the dBm units means that the reference power level is 1mW), the resulting power is:
$$
\begin{align}
  P_s &amp;= 10^{-46/10} \cdot 1 \mbox{mW} \\
      &amp;\approx 2.5 \times 10^{-5} \cdot 1 \mbox{mW} \\
      &amp;\approx 2.5 \times 10^{-8} \, \mbox{W} \\
      &amp;\approx 2.5 \times 10^{-2} \, \mbox{µW}
\end{align}
$$
Similarly for a measurement of -96dBm, the resulting power is:
$$
\begin{align}
  P_n &amp;= 10^{-96/10} \cdot 1 \mbox{mW} \\
      &amp;\approx 2.5 \times 10^{-10} \cdot 1 \mbox{mW} \\
      &amp;\approx 2.5 \times 10^{-13} \, \mbox{W} \\
      &amp;\approx 2.5 \times 10^{-4}  \, \mbox{nW}
\end{align}
$$

Now for the signal-to-noise ratio (SNR), it is simply defined as the ratio of the signal power to the noise power. So in linear scale, that would be:
$$
\begin{align}
  \mbox{SNR} &amp;= \frac{P_s}{P_n} \\
             &amp;\approx \frac{2.5 \times 10^{-8}}{2.5 \times 10^{-13}} \\
             &amp;\approx 10^{5}
\end{align}
$$
which when expressed in decibels would give us:
$$
\begin{align}
  \mbox{SNR}_{\mbox{dB}} &amp;= 10\log_{10} \left(\frac{P_s}{P_n}\right) \, \mbox{dB} \\
             &amp;= 10\log_{10} \left(10^{5}\right) \, \mbox{dB} \\
             &amp;= 50\mbox{dB}
\end{align}
$$
We could of course also have taken a bit of a shortcut to compute that SNR directly from the power measurements in dBm by using logarithmic identities to get:
$$
\begin{align}
  \mbox{SNR}_{\mbox{dB}} &amp;= 10\log_{10} \left(\frac{P_s}{P_n}\right) \, \mbox{dB} \\
             &amp;= 10\log_{10} \left(\frac{\frac{P_s}{P_0}}{\frac{P_n}{P_0}}\right) \, \mbox{dB} \\
             &amp;= 10\log_{10} \left(\frac{P_s}{P_0}\right) - 10\log_{10}\left(\frac{P_n}{P_0}\right) \, \mbox{dB} \\
             &amp;= \left(-46\,\mbox{dBm}\right) - \left(-96\,\mbox{dBm}\right) \\
             &amp;= 50\mbox{dB}
\end{align}
$$