If the Earth were the size of a pool ball, it would meet the regulation smoothness (and even sphericality$^{1}$) criteria. So it's already pretty damn flat.
So to do this we need to assume perfection. Wikipedia gives an average radius as $6371\;\text{km}$.
The surface area of a sphere is $4\pi r^{2}$, giving $510,064,472\;\text{km}^{2}$.
Total water on Earth is $1,386,000,000\;\text{km}^{3}$
Apparently, the depth of a volume spread over an area is simply $\frac{Volume}{Area}$.
Giving:
$\frac{1,386,000,000\;\text{km}^{3}}{510,064,472\;\text{km}^{2}} \approx 2.7\;\text{km}$.
So, the Earth would apparently be covered to a depth of around $2.5\;\text{km}$ (some water would still be in the atmosphere I assume).
This seems pretty deep, but a quick Google reveals other people coming up with the same answer so I guess it's right!
$^{1}$ The xkcd linked by James talks about this report, which concludes that the Earth meets smoothness criteria, but not roundness criteria, as the difference in diameter at the poles is greater than allowed for discrepancies.
The link I used divides the difference in diameter by two and so concludes that as the difference each side is less than the allowable error, it DOES meet the criteria.
It's a bit ambiguous; draw your own conclusion. I don't think it affects my answer particularly, given that it's based on completely idealised calculations anyway.