A sphere of maximum volume is 'cut out' from a solid hemisphere of radius r. The ratio of the volume of the hemisphere to that of the cut-out sphere is:

A solid hemisphere has radius $r$. The largest sphere that can fit inside must touch both the flat base and the curved surface.

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Let the cut-out sphere have radius $R$.

The center of the sphere sits at height $R$ above the flat base.

The top of the sphere is at height $R + R = 2R$ from the base.

For the sphere to touch the curved surface of the hemisphere at the top, this point must lie on the hemisphere's surface.

The hemisphere has height $r$ at its center.

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For the sphere to be tangent to the curved surface:

$2R = r$

$R = \frac{r}{2}$

The cut-out sphere has radius $\frac{r}{2}$.

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Volume of hemisphere:

$V_{\text{hemisphere}} = \frac{2}{3}\pi r^3$

Volume of cut-out sphere:

$V_{\text{sphere}} = \frac{4}{3}\pi R^3$

$V_{\text{sphere}} = \frac{4}{3}\pi \left(\frac{r}{2}\right)^3$

$V_{\text{sphere}} = \frac{4}{3}\pi \times \frac{r^3}{8}$

$V_{\text{sphere}} = \frac{1}{6}\pi r^3$

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The ratio of volumes:

$\text{Ratio} = \frac{V_{\text{hemisphere}}}{V_{\text{sphere}}}$

$= \frac{\frac{2}{3}\pi r^3}{\frac{1}{6}\pi r^3}$

$= \frac{2}{3} \times \frac{6}{1}$

$= \frac{12}{3}$

$= 4$

Therefore, the ratio is $4 : 1$.

The answer is option 4.