A cylinder is open at both ends and is made of 1.5-cm-thick metal.

Thickness of the cylinder $=1.5 \mathrm{~cm}$

External diameter $=12 \mathrm{~cm}$

i.e., radius $=6 \mathrm{~cm}$

also, internal radius $=4.5 \mathrm{~cm}$

Height $=84 \mathrm{~cm}$

Now, we have the following:

Total volume $=\pi \mathrm{r}^{2} \mathrm{~h}=\frac{22}{7} \times 6 \times 6 \times 84=9504 \mathrm{~cm}^{3}$

Inner volume $=\pi \mathrm{r}^{2} \mathrm{~h}=\frac{22}{7} \times 4.5 \times 4.5 \times 84=5346 \mathrm{~cm}^{3}$

Now, volume of the metal $=\operatorname{total}$ volume $-$ inner volume $=9504-5346=4158 \mathrm{~cm}^{3}$\

$\therefore$ Weight of iron $=4158 \times 7.5=31185 \mathrm{~g}=31.185 \mathrm{~kg}\left[\right.$ Given: $\left.1 \mathrm{~cm}^{3}=7.5 \mathrm{~g}\right]$