The balloon has a volume [tex]V[/tex] dependent on its radius [tex]r[/tex]:
[tex]V(r)=\dfrac43\pi r^3[/tex]
Differentiating with respect to time [tex]t[/tex] gives
[tex]\dfrac{\mathrm dV}{\mathrm dt}=4\pi r^2\dfrac{\mathrm dr}{\mathrm dt}[/tex]
If the volume is increasing at a rate of 10 cubic m/s, then at the moment the radius is 3 m, it is increasing at a rate of
[tex]10\dfrac{\mathrm m^3}{\mathrm s}=4\pi (3\,\mathrm m)^2\dfrac{\mathrm dr}{\mathrm dt}\implies\dfrac{\mathrm dr}{\mathrm dt}=\dfrac5{18\pi}\dfrac{\rm m}{\rm s}[/tex]
The surface area of the balloon is
[tex]S(r)=4\pi r^2[/tex]
and differentiating gives
[tex]\dfrac{\mathrm dS}{\mathrm dt}=8\pi r\dfrac{\mathrm dr}{\mathrm dt}[/tex]
so that at the moment the radius is 3 m, its area is increasing at a rate of
[tex]\dfrac{\mathrm dS}{\mathrm dt}=8\pi(3\,\mathrm m)\left(\dfrac5{18\pi}\dfrac{\rm m}{\rm s}\right)=\dfrac{20}3\dfrac{\mathrm m^2}{\rm s}[/tex]
