How can I calculate the pH of the solution created by reacting sodium with water?
You answer is right, but your reasoning is not.

Amount of $\ce{Na} = \frac{0.023}{23} = 0.001$ mole

$[\ce{NaOH}]= \frac{0.001}{0.100} (\mathrm{moleL^-1}) = 0.01$ M

$\downarrow$

$p\ce{OH} = 2 \to p\ce{H} = 14 - 2 = 12$