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 $\dfrac{\mathrm dA}{\mathrm dt}$. $\dfrac{\mathrm dA}{\mathrm dt}=8\pi \times15\times\dfrac5{15^2\pi}=\dfrac{40}{15}=\dfrac83$