Any three positive integers that multiply to make $2009$ would create viable cuboids.
The prime factors of $2009$ are $7\times 7\times 41$, so the options are:
$1 \times1 \times 2009$
$1\times 7\times 287$
$1\times 41\times 49$
$7\times 7 \times 41$
The first three cuboids all have two faces which each require $2009$ stickers ($1\times2009$, $7\times287$ and $41\times49$ respectively) so Ruth cannot cover them.
The last cuboid has surface area: $2\times( 7\times7+7\times41 + 41\times 7) = 1246$
This leaves $2009-1246=763$ stickers left over.
This problem is taken from the UKMT Mathematical Challenges.