A spherical balloon's volume increases by 18p%. What is the percentage change in its radius?

Let the initial volume of the spherical balloon be $V_0$ and its initial radius be $r_0$.
Let the final volume be $V_1$ and its final radius be $r_1$.
The formula for the volume of a sphere is $V = \frac{4}{3}\pi r^3$.
So, we have:

$$V_0 = \frac{4}{3}\pi r_0^3$$ $$V_1 = \frac{4}{3}\pi r_1^3$$

The volume increases by $18p\%$. This means the new volume is the original volume plus $18p\%$ of the original volume.

$$V_1 = V_0 + \frac{18p}{100}V_0$$ $$V_1 = V_0 \left(1 + \frac{18p}{100}\right)$$

Divide both sides of the equation by $\frac{4}{3}\pi$:

$$r_1^3 = r_0^3 \left(1 + \frac{18p}{100}\right)$$

Take the cube root of both sides:

$$r_1 = \sqrt[3]{r_0^3 \left(1 + \frac{18p}{100}\right)}$$ $$r_1 = r_0 \left(1 + \frac{18p}{100}\right)^{1/3}$$

The percentage change in radius is given by the formula:

$$\text{Percentage change} = \frac{\text{Change in radius}}{\text{Original radius}} \times 100\%$$ $$\text{Percentage change} = \frac{r_1 - r_0}{r_0} \times 100\%$$

Substitute the expression for $r_1$ from Step 4:

$$\text{Percentage change} = \frac{r_0 \left(1 + \frac{18p}{100}\right)^{1/3} - r_0}{r_0} \times 100\%$$

Factor out $r_0$ from the numerator:

$$\text{Percentage change} = \frac{r_0 \left[ \left(1 + \frac{18p}{100}\right)^{1/3} - 1 \right]}{r_0} \times 100\%$$ $$\text{Percentage change} = \left[ \left(1 + \frac{18p}{100}\right)^{1/3} - 1 \right] \times 100\%$$

The percentage change in its radius is $\boxed{\left[ \left(1 + \frac{18p}{100}\right)^{1/3} - 1 \right] \times 100\%}$.