Many thanks to Andrei Lazanu 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:

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

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

$2 \times (h \times 2 \times d + h \times 3 \times d + 2 \times d \times 3 \times d)$ = 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°

Therefore angle MBO =30°.

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