Let the length (l), breath (b), and stature (h) be the outside element of an open box and thickness be x.

The volume of metal utilized in box = Volume of outer box – Volume of inner box

Think about outside box,

Length, l = 36 cm

Broadness, b = 25 cm

Tallness, h = 16.5 cm

We realize that the condition of the volume of cuboid is given by,

Volume of cuboid = lbh, where, l, b and h are the length, expansiveness and stature of tank separately

Volume of outer box \[=\text{ }36\left( 25 \right)\left( 16.5 \right)\text{ }=\text{ }14850\text{ }cm3\]

Since the crate is open from top,

Think about inside box,

The thickness of different sides is diminished as follows,

Length, l’ = Length of outer box – 2(thickness of box) \[=\text{ }36\text{ }\text{ }2\left( 1.5 \right)\text{ }=\text{ }33\text{ }cm\]

Expansiveness, b’ = Breadth of outside box – 2(thickness of box)\[=\text{ }25\text{ }\text{ }2\left( 1.5 \right)\text{ }=\text{ }22\text{ }cm\]

Stature, h’ = Height of outside box – thickness of box \[=\text{ }16.5\text{ }\text{ }1.5\text{ }=\text{ }15\text{ }cm\]

Volume of inside box = 33(22)(15) = 10890

Furthermore,

Volume of metal in box \[=\text{ }14850\text{ }\text{ }10890\text{ }=\text{ }3960\text{ }cm3\]

We realize that,

1 cm3 weighs 7.5 g

Thus, 3960 cm3 weighs 3960(7.5) = 29,700 g

Consequently, the heaviness of box is 29,700 g for example 29.7 kg