Answer: \(\dfrac{\pi}{8}\text{ cm}^3\text{ s}^{-1}\).

The volume is

\(V=\frac43\pi r^3.\)

Differentiate with respect to \(r\):

\(\frac{dV}{dr}=4\pi r^2.\)

Also

\(\frac{dr}{dt}=0.5.\)

Using the chain rule,

\(\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}.\)

So

\(\frac{dV}{dt}=4\pi r^2(0.5)=2\pi r^2.\)

When \(r=0.25=\frac14\),

\(\frac{dV}{dt}=2\pi\left(\frac14\right)^2 =2\pi\cdot\frac1{16} =\frac{\pi}{8}.\)

Therefore the volume is increasing at

\(\frac{\pi}{8}\text{ cm}^3\text{ s}^{-1}.\)