Helium is pumped into a spherical balloon at $\displaystyle 2 \frac{ft^3}{s}$. How fast is the radius increasing after 2 minutes?

I've searched some of the other related rates probs posted on this forum, so I apologize if a question like this has been brought up already.

Basically, I understand I'm looking for $\displaystyle \frac{dV}{dt}$ , and that $\displaystyle \frac{dV}{dt} = \frac {dV}{dr} * \frac {dr}{dt} $.

Now, the rate is 2 $\displaystyle \frac{ft^3}{s}$ and the volume of a sphere is $\displaystyle V = \frac{4}{3} * {PI}r^3 $.

Implicitly differentiating V with respect to T gets $\displaystyle \frac {dV}{dt} = 4 * {PI}r^2 \frac{dr}{dt} $

I know that the rate the volume increases with respect to time is $\displaystyle 2 \frac{ft^3}{s}$, and I'm not sure what to do with my 2 minutes (or 120 seconds). Is the volume with respect to time 120 seconds and $\displaystyle 2 \frac{ft^3}{s}$ is the rate with respect to time? That would make more sense I think because the dimensions of cubic feet with a factor of 2 is like a rate and seconds is most definitely a part of time.

However, when I try to solve $\displaystyle 120 = 4PIr^2*2 $ or $\displaystyle \frac{dV}{dt} = \frac {dV}{dr} * \frac {dr}{dt} $ and that's incorrect. Since I don't know the volume with respect to time, how can I properly go about solving for the rate with respect to time?

I get $\displaystyle r = sqrt(\frac{120}{8PI}) $