Answer:
the surface area will increase 16  feet per minute
Step-by-step explanation:
Allow me to revise your question for a better understanding. 
<em>A spherical balloon is inflating with helium at a rate of 64π Start Fraction ft cubed Over min End Fraction ft3 min. How fast is the balloon's radius increasing at the instant the radius is 2 ft? How fast is the surface area increasing? </em>
My answer:  
Given that: 
 (1)
  (1) 
As we know that, the volume of a  spherical balloon can be calculated by using the following formula: 
 where r is the radius
 where r is the radius 
Substitute V into (1), we have: 

In this situation, the instant radius is: 2f 
So we have: 

The radius of the balloon is increasing at a rate of 4 feet per minute 
The the surface area has the formula as: 
SA1 = 4π 
 
If the radius of the balloon is increasing at a rate of 4 feet per minute, so the 
the surface area will increase 16  feet per minute  because: 
SA2 : 4π = 16 (4π
  = 16 (4π ) = 16 SA1
 ) = 16 SA1 
Hope it will find you well.