A solid rectangular piece of iron measures 6 m by 6 cm by 2 cm.

The dimensions of the an iron piece is $6 \mathrm{~m} \times 6 \mathrm{~cm} \times 2 \mathrm{~cm}$, i.e., $600 \mathrm{~cm} \times 6 \mathrm{~cm} \times 2 \mathrm{~cm}(\because 1 \mathrm{~m}=100 \mathrm{~cm}) .$

Its volume $=600 \times 6 \times 2=7200 \mathrm{~cm}^{3}$

Now, $1 \mathrm{~cm}^{3}=8 \mathrm{gm}$

i.e., $7200 \mathrm{~cm}^{3}=7200 \times 8 \mathrm{gm}=57600 \mathrm{gm}$

$\therefore$ Weight of the iron piece $=57600 \mathrm{gm}$

$=57600 \times \frac{1}{1000} \mathrm{~kg} \quad(\because 1 \mathrm{Kg}=1000 \mathrm{gm})$

$=57.6 \mathrm{~kg}$