Radius of the original steel ball

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

$523.6 = \frac{4}{3}\pi r^3$

$r = 5 \, \text{ cm}$ *answer*

Volume of the hollow steel ball is equal to the volume of the original steel ball. Let R = outer radius of the hollow steel ball.

$\frac{4}{3}\pi R^3 - \frac{4}{3}\pi r^3 = \frac{4}{3}\pi r^3$

$\frac{4}{3}\pi R^3 = \frac{8}{3}\pi r^3$

$R^3 = 2r^3$

$R^3 = 2(5^3)$

$R = 6.3 \, \text{ cm}$ *answer*

Thickness of the hollow steel ball

$t = R – r = 6.3 – 5$

$t = 1.3 \, \text{ cm}$ *answer*