Assume the radius of the big ball be $xcm$

The radius of the small ball $=x/4cm$

Let the number of balls $=n$

Then according to the question, we have

Volume of n small balls $=$ Volume of the big ball

$n\times 4/3\pi {{\left( x/4 \right)}^{3}}=4/3\pi {{x}^{3}}$

$n\times \left( {{x}^{3}}/64 \right)={{x}^{3}}$

$n=64$

So, the number of small balls $=64$

Now,

Surface area of all small balls/ surface area of big ball $=64\times 4\pi {{\left( x/4 \right)}^{2}}/4\pi {{\left( x \right)}^{2}}$

$=64/16=4/1$

Thus, the ratio of the surface area of the small balls to that of the original ball is $4:1$