The surface area of the balloon is \(A=4\pi r^2\).
Differentiate both sides with respect to an arbitrary variable representing time:
\(\dfrac{\mathrm dA}{\mathrm dt}=8\pi r\dfrac{\mathrm dr}{\mathrm dt}\)
Meanwhile, the volume of the balloon is \(V=\dfrac43\pi r^3\).
Differentiating with respect to \(t\) yields
\(\dfrac{\mathrm dV}{\mathrm dt}=4\pi r^2\dfrac{\mathrm dr}{\mathrm dt}\)
You're told that, at the point when the radius \(r=15\), the volume of the balloon increases at a rate of \(20\text{ ft}^3/\text{min}\), which means \(\dfrac{\mathrm dV}{\mathrm dt}=20\). Use this to solve for the rate of change of the radius, \(\dfrac{\mathrm dr}{\mathrm dt}\).
\(20=4\pi \times15^2\dfrac{\mathrm dr}{\mathrm dt}\implies\dfrac{\mathrm dr}{\mathrm dt}=\dfrac5{15^2\pi}\)
Substitute this into the equation for the rate of change of the surface area and solve for \(\dfrac{\mathrm dA}{\mathrm dt}\).
\(\dfrac{\mathrm dA}{\mathrm dt}=8\pi \times15\times\dfrac5{15^2\pi}=\dfrac{40}{15}=\dfrac83\)
