The near and far faces appear to be trapezoids, with bases 12 in and 8 in and height 8 in. Hence each of them has an area
.. Atrapezoid = (1/2)(b1 +b2)h
.. = (1/2)(12 in + 8 in)*(8 in)
.. = 80 in^2
The remaining faces unroll into a rectangle 8 in wide and (12 +8 +8+9) in long. The area of a rectangle is the product of its length and width.
.. Arectangle = length*width
.. = (37 in)*(8 in) = 296 in^2
The total surface area is the sum of the rectangle area and the two trapezoidal faces.
.. Atotal = Arectangle + 2*Atrapezoid
.. = 296 in^2 +2*80 in^2
.. = 456 in^2
The 1st selection is appropriate.