Speed Limits on a Cube

The bug will have to go through at least $2$ faces. It is quite apparent that he should go through exactly $2$ and not more to reach the minimum amount of time spent.

Therefore, he'll go through $2$ faces. He can't go through $2$ faces that force a $3$ units/min speed, since they're on the opposite sides. So the best way for him to get to the opposite vertex is to go through the $3$ and $2$ units\min faces.

The following diagram shows the $2$ faces he'll go through, and they're incidentally the same faces the bug goes through on the Daniel's image (the faces have the same colors too).

This is a diagram.

The minimum of the following function is equal to the minimum amount of time the bug spends by going through the cube:

$\displaystyle \frac{\sqrt{2014^2+x^2}}{3}+\frac{\sqrt{2014^2+(2014-x)^2}}{2}\ge\approx 1867.14\approx M$

And this minimum can be reached when $x\approx 1255.12$ .

Wolfram Alpha helped me out a bit here.

Hence, the answer is $\lfloor M \rfloor\equiv \boxed {867}\pmod{1000}$