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:
$1 \times 6$ cups
and
$2 \times 3$ 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 $1 \times 6$ 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 $2 \times3$ cups box is:
$2 \times(h \times2 \times d + h \times3 \times d + 2 \times d
\times3 \times h) = 157250$ 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 $2 \times BM + 2 \times85$ 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 $MBO = 30^{\circ}$
$$\eqalign{ BM &= \frac{85}{2}\tan60^{\circ} \\ \; &=
\frac{85}{2}\sqrt{3} \\ \; &= \frac{85\sqrt{3}}{2}\\
\mbox{The side of triangle ABC is} &= 2 \times d + 2 \times
BM \\ \; &= 2 \times85 + 2 \times\frac{85\sqrt{3}}{2} \\ \;
&= 85(2 + \sqrt{3}) \\ \mbox{The altitude of the equilateral
triangle ABC is} &= AB \times\sin60^{\circ} \\ \; &= 85(2
+ \sqrt{3}) \times\frac{\sqrt{3}}{2} \\ \; &=
\frac{85\sqrt{3}(2 + \sqrt{3})}{2} \\ \mbox{Area of the two
equilateral triangles} &= 85(2 + \sqrt{3})
\times\frac{85\sqrt{3}(2 + \sqrt{3})}{2} \\ \; &= 87149 \;
\mbox{mm}^2 \\ \mbox{Area of the three faces} &= 3 \times83
\times85(2 + \sqrt{3}) \\ \; &= 78989 \; \mbox{mm}^2 \\
\mbox{The total surface area is } &= 166138 \; \mbox{mm}^2}$$
So the best box is a box of 2 x 3 cups.