A=2πRh+2πR² is the total area of the can including the circular ends, where h is the height.
The volume is πR²h=250cc, so h=250/(πR²).
Therefore A in terms of r is given by:
A=(2πR)(250/(πR²)+2πR²=500/R+2πR².
We can differentiate this:
dA/dR=-500/R²+4πR=0 at an extremum.
So, -500+4πR³=0, and R³=500/(4π)=125/π.
And R=5/∛π=3.4139cm approx.
(Substitute this value for R and A=219.6887 sq cm, whereas if we take a value of R slightly more or less than 3.4139cm, we get a value for A bigger than 219.6887 sq cm, so R=3.4139cm gives minimum surface area.)
Note that h=2R for this minimum condition and that a cross-section of the can through the diameters of the top and bottom form a perfect square, so this question could have been solved without calculus, because volume=250=2πR³.