You need to calculate the change in the moment of inertia of the Earth and use conservation of angular momentum. In the case of eg. a large dam, (most of) the water will ultimately come from the sea, effectively removing a thin layer of water. The contribution of the removed water to the moment of inertia depends on the distance from the rotation axis and hence on the latitude. This is a fairly simple calculation if we assume the world is all ocean. A more precise calculation would have to take into account the shapes of the oceans. You also need the moment of inertia of the earth, which depends on the density as a function of depth.
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 of the earth is then
$$\frac{mR^2(\cos^2 L - \frac{2}{3})}{I}$$
(where $I$ is the Earth's moment of inertia, $8.04×10^{37}$ kg·m$^2$) or
$$1.97×10^{-11}(\cos^2 L - \frac{2}{3}) .$$
Multiplying by the number of microseconds in a day ($8.64 \times 10^{10}$) gives:
$$1.7(\cos^2 L - \frac{2}{3}) = 1.7 × 0.071 = 0.12\,\mu\text{s}$$
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 (fairly 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. I presume NASA took into account the shapes of the oceans and got a more accurate value for this term.
