A hemispherical bowl is made of steel 0.25 cm thick.

Inner radius = 5 cm

Outer radius = 5 + 0.25 = 5.25

Volume of steel used = Outer volume-Inner volume

$=2 / 3 \times \pi \times\left((5.25)^{3}-(5)^{3}\right)$

$=2 / 3 \times 22 / 7 \times\left((5.25)^{3}-(5)^{3}\right)$

$=41.282 \mathrm{~cm}^{3}$