Twenty seven solid iron spheres,

Solution:

(i)Radius of 1 solid iron sphere $=r$

Volume of 1 solid iron sphere $=\frac{4}{3} \pi r^{3}$

Volume of 27 solid iron spheres $=27 \times \frac{4}{3} \pi r^{3}$

27 solid iron spheres are melted to form 1 iron sphere. Therefore, the volume of this iron sphere will be equal to the volume of 27 solid iron spheres. Let the radius of this new sphere be $r$ '.

Volume of new solid iron sphere $=\frac{4}{3} \pi r^{3}$

$\frac{4}{3} \pi r^{\prime 3}=27 \times \frac{4}{3} \pi r^{3}$

$r^{\prime 3}=27 r^{3}$

$r^{\prime}=3 r^{\prime}$

(ii) Surface area of 1 solid iron sphere of radius $r=4 \pi r^{2}$

Surface area of iron sphere of radius $r^{\prime}=4 \pi\left(r^{\prime}\right)^{2}$

$=4 \pi(3 r)^{2}=36 \pi r^{2}$

$\frac{\mathrm{S}}{\mathrm{S}^{\prime}}=\frac{4 \pi r^{2}}{36 \pi \mathrm{r}^{2}}=\frac{1}{9}=1: 9$