Let radius of cylinder = R
and, radius of each sphere = r

Height of cylinder, H = 2*diameter of cylinder = 2*2R = 4R
Volume of cylinder = \(\pi*R^2H\) = \(\pi*R^2*4R\) = \(4\pi*R^3\)
Total surface area of cylinder = \(2\pi*R^2+2\pi*RH\) = \(10\pi*R^2\)

Since three spheres are made out of the cylinder,
3*Vol. (each sphere) = Vol. (cylinder)
\(3*\frac{4}{3}\pi*r^3\) = \(4\pi*R^3\)
Simplifying, we get, r = R

Surface area of each sphere = \(4\pi*r^2\) = \(4\pi*R^2\)
Surface area of all three spheres = \(3*4\pi*R^2\) = \(12\pi*R^2\)

Percentage difference in surface area = \(\frac{Area(All Spheres)-Area(Cylinder)}{Area(Cylinder)}*100\)% = \(\frac{12\pi*R^2-10\pi*R^2}{10\pi*R^2}*100\)% = \(\frac{2}{10}*100\)% = 20%

Ans is D IMO