An iron spherical ball has been melted and recast into smaller balls of equal size.

Solution:

Let the radius of the big metallic ball is 4r. Therefore, the volume of the big metallic ball is

$V=\frac{4}{3} \pi \times(4 r)^{3}$

The metallic sphere is melted to produce small balls of radius $\frac{4 r}{4}=r$. Then, the volume of each of the small balls is

$V_{1}=\frac{4}{3} \pi \times(r)^{3}$

Since, the volume of the big metallic ball is equal to the sum of the volumes of the small balls, we have the number of produced small balls is

$\frac{V}{V_{1}}=\frac{\frac{4}{3} \pi \times(4 r)^{3}}{\frac{4}{3} \pi \times(r)^{3}}$

$=(4)^{3}$

$=64$

Hence, the number of small balls is 64

The surface area of the big ball is

$S=4 \pi \times(4 r)^{2}$

The surface area of each of the small ball is

$S_{1}=4 \pi \times(r)^{2}$

Therefore, the total surface area of the 64 small balls is

$S_{2}=64 \times 4 \pi \times(r)^{2}$

Now, we compute the following ratio

$\frac{S_{2}}{S}=\frac{64 \times 4 \pi \times(r)^{2}}{4 \pi \times(4 r)^{2}}=4$

$\Rightarrow S_{2}=4 S$

Hence, the total surface area of the small balls is equal to four times the surface area of the original big ball.