This question is a bit unclear, still I will try to give some sort of answer.

Some intial remarks:

First, at the moment noone suceed in proving primality for a 100 million (decimal) digit number. The current record is (I believe) close to 13 million digits (in binary this would still not be 100 million).

Second, one does not do these test by trial division as your description seems to suggest.

Having said this, as a thought experiment a very rough and overly optimistic calculation:

Suppose you have a 100 million digit number. Then you will have to test whether it is divisibile by numbers of size up to its square root, that is 50 million digit numbers.
So, you test $10^{50 000 000}$ numbers.

Suppose you do your divison with only one processor instruction. I am not overly knowledgeable on processor speeds but according to the list here let's assume you do 50 GIPS so $5*10^9$ instruction per second.

Now, in an hour you will do, let's be generous, $2*10^{12}$ instructions, by our assumption divisoions.

So you need $10^{50 000 000}/(2*10^{12}) = 5 * 10^{49 999 987}$ processors.

As said, you cannot do this like this. Also note that using the approach you sketch you would effectively find a factor and thus a factorization. For factoring the current records are way smaller then the ones for primes I mentioned above. It is a major challenge to factor numbers with (low) hundreds of digits. Note that that current RSA-keys are of size a a thousand or two (maybe four) thousand bits. So (higher) hundreds to a thousand decimal digits only.

P.S. Towards the end of my writing, I saw paul garrett's comment which is similar. Perhaps the details are useful. And, sorry to those who mind, for answering the off-topic question.