Note: Question as originally posted asked about a photodiode, not a solar cell.
If you are not designing the device yourself, you don't need to know all those values. Most photodiodes you buy will simply have a dark current spec. If you need to know how this varies with reverse bias, you can work out Is from the dark current and the measurement conditions (V and T).
For example, Fairchild QSD2030 specifies a dark current of 10 nA with 10 V reverse bias at 25 C. Given the specified reverse bias is so high, we know this dark current is essentially equal to Is, so we know Is ~= 10 nA.
Edit
Okay, so you really want to get those values as an academic excercise.
A is just the area of the device. It can be whatever you design it to be.
Dn, Dp are the diffusion coefficients. You can get them from your textbook or from the Ioffe Institute
τn and τp are the carrier lifetimes. You can also find these values at the Ioffe Institute site.
Nd and Na are the donor and acceptor doping concentrations. You could design these however you like. Numbers between 1015 and 1020 are at least soft of reasonable.
The product NcNv is equal to ni2, the square of the intrinsic carrier concentration. You can also get this value from the Ioffe Institute on another page.
How can a charge neutral substance have a potential?
To say that a PN junction has built-in potential isn't to say that the PN junction has a potential relative to ground or infinity etc.
How can there be an inherent potential in a doped semi-conductor if it
is charge neutral?
Charge has been separated within the PN junction and, thus, there is an electric field across the depletion region and an associated potential difference. A charged capacitor is neutral but there is a potential difference (voltage) across the dielectric. There are some similarities but...
However, I don't understand why we can't measure the voltage drop
across the PN block with a voltmeter.
As explained, for example, here, the built-in potential is not readily measured with, e.g., a voltmeter. In other words, this question has been asked here several times (which means you are not the only one perplexed by this - most are at first) and there are good answers already available.
Best Answer
Silicon has a density of \$2.33\ {\rm g/cm^3}\$. And its atomic weight is about 28. Knowing that 1 atomic mass unit is about \$1.66\times10^{-24}\ {\rm g}\$, we can find that there are about \$5\times10^{22}\$ silicon atoms per cubic centimeter of material.
If the doping density in a solar cell is about \$10^{17}\ {\rm cm^{-3}}\$ (I don't know if this is the most likely value, but it's a common value for fairly strongly doped material in other devices and the first one I came up with in a google search for "solar cell dopant concentration"), then the fraction of silicon atoms replaced by dopants in that material is about \$2\times10^{-6}\$, or 2 ppm.
Of course if you have a more accurate figure for the doping concentration in some particular solar cell you're interested in, you can easily calculate the value for that device yourself.