Iron Balls

Since they are similar,

$27(r^3)=R^3$

$27(2^3)=R^3$

$216=R^3$

$R=6$

The real problem is knowing if the balls are hollow or not. In this problem, these balls are completely made of iron. Which theorically means that when melted, the liquid iron will equal the volume of the ball.

{4}/{3} *pi * r^{3} =Sphere volume

Each ball has radius 2 so

{4}/{3} *pi * 8 = Small ball volume

There are 27 balls melted so

{4}/{3} * pi * 8 * (27) = Big ball volume

Using some algebra

{4}/{3} * pi * 8* (27) = Big ball volume

{4}/{3} * pi * r^{3} = Big ball volume

Then,

r^{3} = 8*27

r^{3}= 2^3 * 3^3

r^{3}= 6^3

Therefore, r=6