Answer:
58 in^2
Explanation:
The area of one face of the 3" cube is (3 in)^2 = 9 in^2. Then the 6 faces of that will have an area of ...
6·9 in^2 = 54 in^2
The area of the top of the paperweight is unaffected by the area of the added cube: the area the smaller cube covers is exactly the same as its exposed top area. So, the net effect of the added 1" cube is to increase the total area by that cube's lateral area: 4 faces at 1 in^2 for each face, a total of 4 in^2.
Then the total surface area of the paperweight is ...
54 in^2 + 4 in^2 = 58 in^2