[tex]V=\dfrac43\pi r^3[/tex]
[tex]\dfrac{\mathrm dV}{\mathrm dt}=4\pi r^2\dfrac{\mathrm dr}{\mathrm dt}[/tex]
You're given that the volume of the balloon is increasing at a rate of 12 cm^3/sec, which means [tex]\dfrac{\mathrm dV}{\mathrm dt}=12[/tex]. The balloon pops when [tex]r=5[/tex], so you have enough information to find the rate of change of the radius at this time:
[tex]12=4\pi(5)^2\dfrac{\mathrm dr}{\mathrm dt}\implies \dfrac{\mathrm dr}{\mathrm dt}=\dfrac3{25\pi}\approx0.0382[/tex] cm/sec
