Find the volume of iron required to make an open box whose external dimensions are 36 cm × 25 cm × 16.5 cm,

External length $=36 \mathrm{~cm}$

External width $=25 \mathrm{~cm}$

External height $=16.5 \mathrm{~cm}$

External volume of the box $=36 \times 25 \times 16.5=14850 \mathrm{~cm}^{3}$

Thickness of iron $=1.5 \mathrm{~cm}$

$\therefore$ Internal length $=36-(1.5 \times 2)=33 \mathrm{~cm}$

Internal width $=25-(1.5 \times 2)=22 \mathrm{~cm}$

Internal height $=16.5-1.5=15 \mathrm{~cm}$ (as the box is open)

Internal volume of the box $=33 \times 22 \times 15=10890 \mathrm{~cm}^{3}$

Volume of iron $=$ External volume $-$ Internal volume $=14850-10890=3960 \mathrm{~cm}^{3}$

Given:

$1 \mathrm{~cm}^{3}$ of iron $=8.5$ grams

Total weight of the box $=3960 \times 8.5=33660$ grams $=33.66$ kilograms