Why does a right-circular cylinder helps reduce surface area of the former International Prototype of Kilograms

Though it's not precisely clear, it's possible that they meant something like this:

The surface area of a cylinder of unit volume is minimized when its height is equal to its diameter.

So it wasn't claiming that a cylinder was the minimum-surface-area shape, but rather that a cylinder of those particular dimensions had a lower surface area than cylinders of the same volume with different dimensions.

This statement can be proved fairly straightforwardly: a cylinder of diameter $D$ and height $H$ has volume $V=frac{1}{4}pi D^2H$ and surface area $S=frac{1}{2}pi D^2+pi DH$. If we fix the volume to be a constant $V_0$, then we have that $H=frac{4V_0}{pi D^2}$. Substituting, we get the surface area as a function of the diameter:

$$S=frac{1}{2}pi D^2+frac{4V_0}{D}$$

To minimize $S$, we first find points where the derivative with respect to $D$ is zero:

$$frac{dS}{dD}=pi D-frac{4V_0}{D^2}=0$$

The solution to this is $D=(4V_0/pi)^{1/3}$. To see whether this is a maximum or a minimum, we can find the second derivative at that point:

$$frac{d^2S}{dD^2}=pi+frac{8V_0}{D^3}$$

This is always positive for $D>0$, so this is indeed a minimum.

So the diameter at which surface area is minimized is $D=(4V_0/pi)^{1/3}$. This means that the height at which surface area is minimized, for a constant volume, is:

$$H=frac{4V_0}{pi(4V_0/pi)^{2/3}}=(4V_0/pi)^{1/3}=D$$