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Mark the center O of the regular hexagon at the base and connect the center with two consecutive vertices A and B of the hexagon.
You will get an equilateral triangle AOB with the side length of 2 inches.
You can place 4 equilateral triangles with the side length of 1 unit inside the triangle AOC.
Next, take into account that the hexagonal base comprise of 6 such triangles congruent to triangle AOB.
It means that you can place 6*4 = 24 small triangles inside the hexagon base.
You then multiply this number, 24, by 2 to account for the height of the package and the height of the candy bar.
So your answer is 24*2 = 48 candy bars in one package.
Solved.