The source cited by user140606 is incorrect. This is my Answer to a similar Question on earthscience.SE :
You need to calculate the change in the moment of inertia of the Earth and use conservation of angular momentum (the rotation period is proportional to the moment of inertia). Most of the water will ultimately come from the oceans, effectively removing a thin layer of water.
Jerry Mitrovica discusses this effect (in reverse) in a Nautilus interview:
Is water moving off glaciers, slowing the Earth’s rotation, this time analogous to a figure skater putting arms out?
Right. Glaciers are mostly near the axis. They’re near the North and South Poles and the bulk of the ocean is not. In other words, you’re taking glaciers from high latitudes like Alaska and Patagonia, you’re melting them, they distribute around the globe, but in general, that’s like a mass flux toward the equator because you’re taking material from the poles and you’re moving it into the oceans. That tends to move material closer to the equator than it once was.
So the melting mountain glaciers and polar caps are moving bulk toward the equator?
Yes. Of course, there is ocean everywhere, but if you’re moving the ice from a high latitude and you’re sticking it over oceans, in effect, you’re adding to mass in the equator and you’re taking mass away from the polar areas and that’s going to slow the earth down.
The contribution of the removed water to the moment of inertia depends on the distance from the axis and hence on the latitude. This is a simple calculation if we assume the world is all ocean.
The moment of inertia of the lake is $m(R\cos L)^2$ and the moment of inertia of a spherical shell is $\frac{2}{3}mR^2$, where m is the mass of water, $R$ is the Earth's radius and $L$ is the latitude of the lake ($30.82305$ degrees for Three Gorges). The relative change in the moment of inertia $I$ of the Earth is then
$$\frac{mR^2}{I}(\cos^2 L - \frac{2}{3})
= \frac{39 \times 10^{12} \times(6.37 \times 10^{6})^2}{8.04×10^{37}}(\cos^2 L - \frac{2}{3})$$
$$=1.97×10^{-11}(\cos^2 L - \frac{2}{3})$$
Multiplying by the number of microseconds in a day ($8.64 \times 10^{10}$) gives
$$1.7(\cos^2 L - \frac{2}{3}) = 1.7 × 0.071 = 0.12$$
microseconds.
Why the difference from NASA's $0.06$ ? Note that the expression changes sign at $\cos^2 L = \frac{2}{3}$ or $L ≈ 35$ degrees (pretty close to the latitude of Three Gorges). The Earth will actually speed up if the lake is at high latitudes and slow down if it is at the equator. The $\frac{2}{3}$ term comes from the "all ocean" assumption.
As I understand this paper, the $\frac{2}{3}$ term should be multiplied by $\frac{1.414}{1.38}$ to account for the shape of the oceans (search in the PDF for those numbers), resulting in $0.09$ microseconds (.
