Explanation:
this might be an intended trick question.
it suggests that the only difference for the surface area are the missing 2 small rectangles at the top and bottom areas.
while that is a difference, yes, we have now also the inner walls of the holes as additional surface area !
let's start with the subtraction of the 2 hole areas :
each cut-out rectangle is 1.5×2 in² = 3 in²
so, both together are 2×3 = 6 in²
and now for the additional inner walls :
first we need to find the dimensions of the cube.
as a cube, all sidelines are of equal length, and each side area is a square.
it has 6 side areas:
294 / 6 = 49 in²
each side area is a square of 49 in².
and that means each side is sqrt(49) = 7 in long.
therefore, the height of the inner walls is also 7 in.
now we know for the inner walls of the hole :
2 are 1.5×7 in² = 10.5 in²
2 are 2×7 in² = 14 in²
that makes the new surface area
294 - 6 + 2×10.5 + 2×14 = 288 + 21 + 28 = 337 in²