It is easy to see that there are an infinite number of prime numbers, so there is no last prime with respect to the usual ordering 2,3,5,7, &c.

BUT..... We can order the prime numbers in alphabetical order according to their name in English (ignoring spaces and hyphens).

So "three" comes before "twenty-three" which comes before "two".

What is the *last* prime number in alphabetical order?

The names of the powers of ten are:

hundred, thousand, million, billion (10**9), trillion, quadrillion, quintillion, ..., decillion, undecellion, dodecillion, tredecillion, quattordecillion, ... vigintillion (10**63)

So the last prime should be close to two 2 x 10**63. (So you will need arbitrary precision math unless you want to use a list of prime numbers from the Internet).

You can use a probalistic primality test like Fermat's test.

Using the number descriptions that you provided, the highest number alphabetically would be:

two vigintillion two undecillion two trillion two thousand two hundred something.

Something would be an odd number between 1 and 99 if, indeed, any of these turn out to be prime.

Average spacing between primes at N is about equal to ln(N). ln(10^63) = 145 so it does seem quite possible that one of them would be prime.