Electrons and Holes in a semiconductor
First, concentrations of electrons and holes must obey the law of mass action (at equilibrium):

$$np=n_i^2$$

Your calculated values clearly violate this for silicon, therefore you can easily tell your calculation is wrong.



It's hard to know what approach an instructor asking this question would want you to follow. If someone asked me to answer this question, I would just write down the answer. If you don't know it off the top of your head, consider what \$n_i\$ represents. If you still don't know the answer or want to just use some basic equations, continue on.



The problem is easier than you are making it. All semiconductors in equilibrium follow the previous law, and also are charge neutral:

$$n + N_A^- = p + N_D^+$$

In an undoped semiconductor the acceptor and donor concentrations are both zero (by definition) and therefore you are left with:

$$n=p$$

Calculating the actual numbers is left as an exercise for the reader. (Hint: If you need a calculator, it's probably wrong)