B

Step 1: Calculate the actual radius of the satellite's orbit which is 1.0 km too large than the required geostationary orbit. Therefore, the radius of the satellite's orbit is given by:
$r_{actual} = r_{geostationary} + 1.0 \text{ km} = 42000 \text{ km} + 1.0 \text{ km} = 42001 \text{ km}$

Step 2: Calculate the orbital period of the satellite using the formula for a satellite's period:
$$T = 2\pi \sqrt{\frac{r^3}{GM}}$$ where G is the gravitational constant and M is the mass of the Earth. However, since we have the geostationary condition T = 24 hours for radius 42000 km, we can find the new rate.
Step 3: The angular velocity of the Earth is given by:
$$\omega_{earth} = \frac{2\pi}{T_{earth}} = \frac{2\pi}{24 \text{ hr}} \approx \frac{2\pi}{86400 \text{ s}} \approx 7.272 \times 10^{-5} \text{ rad/s}$$

Step 4: The satellite’s actual angular velocity will be slightly different but we are interested in the linear speed of the point on the surface directly below the satellite:
$$v = r \omega$$
We consider the radius of the Earth which is 6400 km. The surface point moves at the same speed as the angular velocity of the Earth.
$$v_{below} = R \times \omega_{earth} = 6400 \times 10^{3} \times 7.272 \times 10^{-5} \approx 464 \text{ m/s}$$
Step 5: Since the satellite is above a point that is stationary only if at the height of the geostationary orbit, this point will move in the direction of Earth's rotation (from West to East). Therefore, the point below it will move to the East at a calculated speed.
Therefore, the answer is: 464 m/s to the East.