A human brain is a collection of cells that are massively interconnected. Assuming we knew how to build and program one, how much data would we need to recreate any particular human? That is, what is the best estimates for the total size (in bytes) needed to store a Homo sapiens mindstate? In the following I'm going to try to work out an upper-bound estimate.
Connectivity Information Content
There are about a 100 billion neurons in the human brain. Each one makes, on average, 10,000 connections as inputs. How much information is stored in the connectivity graph of the neural net? How many ways can a single neuron make 10,000 connections? Just
$$ \left(\begin{array}{c} N\\m \end{array}\right) = \left(\frac{N!}{m! (N-m)!}\right) $$
where $N$ is the number of neurons and $m$ the number of connections. Given that each neuron is approximately independent, we then have for the entire brain a total number of states:
$$ \left(\begin{array}{c} N\\m \end{array}\right)^N = \left[\frac{N!}{m! (N-m)!}\right]^N $$
The information content in bits is then
$$\log_2 \left[ \frac{N!}{m!(N-m)!} \right]^N = N \left( N\log_2 N - m\log_2 m - (N-m)\log_2 (N-m) \right) $$
Now the neural system is sparsely connected that is the number of connections, 10,000, is far less than the number of neurons, 100 billion, so $N>>m$. This allows the simplifying assumption
$$ \begin{array}{c} \log_2 (N-m) = \log_2 N \\ \log_2 N_{states} = Nm\log_2\frac{N}{m} \end{array} $$
For the human brain: $ 10^{11} \cdot 10^4 \cdot \log_2 10^7 = 23 \cdot 10^{15} $ bits - That's 3 petabytes! (about 2,000 modern hard drives).
Now, actually the brain is compartmentalized and any given neuron can't make a connection to any of the other 100 billion neurons. Instead, it can only connect to an average subset of $N_0$ neurons. Let's say only a million potential partners are available for any given neuron. The change is:
$$\log_2 N_{\textrm{states}} = Nm\log_2\frac{N_0}{m} = 10^{11} \cdot 10^4 \cdot \log_2 100 = 7 \cdot 10^{15}$$
1 petabyte, not much of a change, since the sheer number of synapses involved sets the petabyte figure.
Synaptic Weight Information Content
The above calculation naively assumes that each connection is binary, just like a junction. In reality, each connection is a synapse that acts like a complex biochemical amplifier of electronic signals that holds lots of state variables. If we assume each synapse has on average σ state variables with r bits of resolution, then we have the additional information source of $ N\cdot m \cdot\sigma\cdot r$.
An oversimplified assumption is that there are only two variables: the receptor number and the size of the synapse, each varying within a 6-bit range (a 64-fold range). This leads to the estimate of 1015 $\cdot$ 2 $\cdot$ 6 = 12 petabits = 1.5 petabytes to store the synaptic strengths in addition to the connectivity diagram.
There are certainly more synaptic state variables than this. We can set an upper limit of perhaps 100 6-bit variables arising from variable local synaptic protein levels and modification states (phosphorylation, ubiquitinylation, etc.). This implies a much larger storage size of
$$ 10^{15} \cdot 100\cdot 6 = 600 \textrm{petabits} = 75 \textrm{petabytes} $$
There is ambiguity at how well such raw data compresses, but it's probably safe to assume a information density of 1-100 petabytes to store a human mindstate.
Note that there are many other variables, such as global protein expression levels and of course the transient electrical state of neurons but these variables tend to scale per-neuron, not per-synapse, and so contribute a much smaller amount of information that we can neglect for an order of magnitude analysis. (i.e. We have less than a gigabyte of genomically coded "human nature".)
This implies that static immortality amounts to several million dollars of storage space, if only we knew how to tease the ghost from the skein and rehaunt another.