Step 1: Recall the basic DNA helix numbers.
In the standard Watson and Crick B-DNA model, one full turn of the helix always contains 10 base pairs, this is a fixed structural constant you can rely on directly.
Step 2: Scale it up to 12 turns.
$12 \times 10 = 120$ base pairs are present across the whole stretch of DNA described in the question.
Step 3: Convert base pairs into individual bases.
Every base pair is made of two bases, one on each strand, so the total number of nitrogenous bases is $120 \times 2 = 240$.
Final answer: Option 3, 240 nitrogenous bases in total.