Answer:
1759 square meters
Explanation:
You want the surface area of a cuboid with a half-cylinder top.
Lateral area
The lateral area of the figure is the product of the length of the mailbox (0.45 m) and the perimeter of the end. The perimeter of the end is the sum of the lengths of the straight sides and half the circumference of a circle with diameter 0.3 m.
P = 0.3 + 2·0.4 + π/2(0.3) = 1.571 . . . . . meters
LA = Ph = (1.571 m)(0.45 m) = 0.70695 m²
End area
The end area is twice the area of the rectangular portion of the end, plus the area of a circle 0.3 m in diameter.
EA = (0.3 m)(0.4 m) + 3.14(0.3/2 m)² = 0.31065 m²
Total area
The total area of 1 mailbox is ...
LA +EA = 0.70695 m² +0.31065 m² = 1.0176 m²
Then the area of 1728 mailboxes is ...
1728 × 1.0176 m² ≈ 1758.4 m² ≈ 1759 m²
About 1759 square meters of aluminum will be needed for the 1728 mailboxes.
__
Additional comment
This presumes there is no waste in cutting the semicircular shape from the supplied aluminum.