Many thanks to Andrei of School 205 Bucharest for the inspiration for this solution. Well done Andrei. One might have expected the triangular solution to be best. I wonder what happens if you chop off the corners?

Using the following notation:

$h$ - for the height of the cup and

$d$ - for the diameter of the cup.

Test each of the two possible forms of the box, one with a rectangular base and one with a triangle as its base:

Cuboidal box

As there are six cups they could be ordered in two ways:

1 x 6 cups

and

2 x 3 cups

For any cuboid box, the surface area is: 2 times the top area + 2 times the side area + 2 times the front area:

The total area of a 1 x 6 cups box is:

$2 x (h x 6 x d + d x 6 x d + d x h) = 185470$ mm $^{2}$

The total area of a 2 x 3 cups box is:

$2 x (h x 2 x d + h x 3 x d + 2 x d x 3 x h) = 157250$ mm $^{2}$

For a triangular prism box This case is illustrated in the figure of the problem.

The base is an equilateral triangle. The surface area is 2 times the area of the equilateral triangle + 3 times the area of each of the faces.

First calculate the side of the equilateral triangle $ABC$ .

This is 2 x $BM$ + 2 x $85$ mm.

$BM$ is a tangent to the circumference of a corner cup (centre $O$ , radius $d / 2 = 85 / 2$ mm).

Also the angle $MOB = 60^{\circ}$

Therefore angle $MBO = 30^{\circ}$

$\begin{matrix} BM & = \frac{85}{2} \tan 60^{\circ} \\ = \frac{85}{2} \sqrt{3} \\ = \frac{85 \sqrt{3}}{2} \\ The side of triangle ABC is & = 2 x d + 2 x BM \\ = 2 x 85 + 2 x \frac{85 \sqrt{3}}{2} \\ = 85 (2 + \sqrt{3}) \\ The altitude of the equilateral triangle ABC is & = AB x \sin 60^{\circ} \\ = 85 (2 + \sqrt{3}) x \frac{\sqrt{3}}{2} \\ = \frac{85 \sqrt{3} (2 + \sqrt{3})}{2} \\ Area of the two equilateral triangles & = 85 (2 + \sqrt{3}) x \frac{85 \sqrt{3} (2 + \sqrt{3})}{2} \\ = 87149 {mm}^{2} \\ Area of the three faces & = 3 x 83 x 85 (2 + \sqrt{3}) \\ = 78989 {mm}^{2} \\ The total surface area is & = 166138 {mm}^{2} \end{matrix}$

So the best box is a box of 2 x 3 cups.