There are two acceptable ways of solving this. I'll show both (just in case one of them doesn't align with the information from your textbook).
Option 1 (fast and easy):
The following is the equation for the area of a regular pentagon with side length s.
[tex]A= \dfrac{s^2}{4} \times \sqrt{5(5+2\sqrt{5})}[/tex]
Plug s=12 into the equation.
[tex]A= \dfrac{12^2}{4} \times \sqrt{5(5+2\sqrt{5})}[/tex]
[tex]\approx 247.7[/tex]
The third choice should be your answer.
Option 2 (much more complicated):
The area of a regular polygon can be solved with the following formula.
[tex]A=\dfrac{\text{perimeter} \times \text{apothem}}{2}[/tex]
Let's solve for the apothem.
A pentagon can be divided into 5 congruent isosceles triangles. The vertex angle will be 360/5, or 72 degrees.
An isosceles triangle can be divided into two right triangles. One of the angles of the right triangle will be 72/2, or 36 degrees.
The leg opposite of the 36 degree angle will be 12/2, or 6 inches long.
[tex]\text{apothem}=\dfrac{6}{\tan{36}}[/tex]
[tex]\approx 8.26[/tex]
The apothem is about 8.26 inches.
The perimeter is 12*5, or 60 inches.
[tex]A=\dfrac{\text{perimeter} \times \text{apothem}}{2}[/tex]
[tex]=\dfrac{60 \times 8.26}{2}[/tex]
[tex]\approx 247.8[/tex]
(it seems that we're off by 0.1 due to rounding errors)
The answer is the third choice. Hope this helps! :)
