The given log weighs \(18.48\) quintals \[ = 1848 \text{ kilograms} \] ∴ Volume of the log \[ = \frac{1848}{3} \text{ cubic decimeters} = 616 \text{ cubic decimeters} \] Let the radius of the log be \( r \) dm and the height be \( h \) dm. ∴ Curved surface area of the log \[ 2\pi rh = 440 \quad \text{(i)} \] And volume of the log \[ \pi r^2 h = 616 \quad \text{(ii)} \] Dividing equation (ii) by equation (i): \[ \frac{\pi r^2 h}{2\pi rh} = \frac{616}{440} \Rightarrow \frac{r}{2} = \frac{616}{440} \Rightarrow r = \frac{616 \times 2}{440} = \frac{28}{10} \] ∴ Diameter of the log \[ = 2r = 2 \times \frac{28}{10} = 5.6 \text{ dm} \]