For a satellite of mass $m$ orbiting Earth at radius $r$ (from the centre), the orbital velocity is:

Set gravitational force equal to centripetal force on the satellite: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ The $m$ cancels. Solve for $v$: $v^2 = \dfrac{GM}{r}$, so $v = \sqrt{\dfrac{GM}{r}}$. Orbital velocity at Earth's surface is about $7.9$ km/s (about 28,000 km/h). Option B looks similar but has a 2 inside the square root — that's the **escape velocity** $v_e = \sqrt{2GM/r}$, which is $\sqrt{2}$ times the orbital speed at the same radius. A textbook trap.