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

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.