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