A polar satellite orbits 800 km above the Earth…

Step 1. Period. $T = 2\pi\sqrt{\dfrac{(7.2 \times 10^{6}\ \text{m})^3}{4.01 \times 10^{14}\ \text{m}^3/\text{s}^2}} = 6062\ \text{s} = 101\ \text{min}$.

Step 2. Orbits a day. $\dfrac{86\,400\ \text{s}}{6062\ \text{s}} = 14.25$. A day on the clock is $86\,400\ \text{s}$. The Earth turns once in $86\,164\ \text{s}$, a little less, and that is the time we need for the shift.

Step 3. Shift. The Earth turns $\dfrac{6062\ \text{s}}{86\,164\ \text{s}} \times 360^\circ = 25.3^\circ$, which at the equator is $\dfrac{25.3}{360} \times 2\pi \times 6400\ \text{km} = 2829\ \text{km}$.