Solution:

The orbital speed \( v \) of a satellite is given by:

\[
v = \sqrt{\frac{GM}{R}}
\]

where \( R \) is the distance from the center of the Earth to the satellite and \( g \) is the acceleration due to gravity at the Earth's surface. Using the relation \( g = \frac{GM}{R_e^2} \), where \( R_e \) is the Earth's radius, we can rewrite the formula as:

\[
v = \sqrt{g \cdot \frac{R_e^2}{R}}
\]

Here, \( R = R_e + h \), where \( h = 0.25 \times 10^6 \) m is the height above the surface of the Earth.

Now, substituting values:

\[
R = 6.38 \times 10^6 + 0.25 \times 10^6 = 6.63 \times 10^6 \, \text{m}
\]

\[
v = \sqrt{9.8 \times \frac{(6.38 \times 10^6)^2}{6.63 \times 10^6}}
\]

Solving:

\[
v \approx 7.76 \, \text{km/s}
\]

So, the orbital speed of the satellite is approximately \( 7.76 \, \text{km/s} \).