If the radius of a sphere is doubled, then the volume of the sphere becomes _____ times the previous volume.

The correct answer is eight.

Let us derive the relationship between the initial volume and the new volume of the sphere step-by-step.

Let the initial radius of the sphere be $r$.

The formula for the volume $V$ of a sphere of radius $r$ is given by:

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

According to the question, the radius of the sphere is doubled. Let the new radius be $R$.

$R=2r$

Now, let us calculate the new volume $\mathrm{V\text{'}}$ using the new radius $R$:

$\mathrm{V\text{'}}=\frac{4}{3}\pi R^3$

Substitute $R=2r$ into the formula for the new volume:

$\mathrm{V\text{'}}=\frac{4}{3}\pi {\left(2r\right)}^3$

$\mathrm{V\text{'}}=\frac{4}{3}\pi \times 8r^3$

$\mathrm{V\text{'}}=8\times \left(\frac{4}{3}\pi r^3\right)$

Since the initial volume is $V=\frac{4}{3}\pi r^3$, we can rewrite the equation as:

$\mathrm{V\text{'}}=8V$

Therefore, when the radius of a sphere is doubled, its volume becomes 8 times the previous volume.