To solve this problem, we need to use the formulas for gravitational potential and gravitational acceleration at a height $h$ above the Earth's surface.$\\$Given:$\\$• Gravitational potential $V = -5.4 \times 10^7 \, \mathrm{J/kg} \\$• Gravitational acceleration $g = 6.0 \, \mathrm{m/s^2} \\$• Radius of Earth $R = 6400 \, \mathrm{km} = 6400 \times 10^3 \, \mathrm{m} \\$The formula for gravitational potential at a height $h$ is:$\\ V = -\frac{GM}{R + h} \\$The formula for gravitational acceleration at a height $h$ is:$\\ g = \frac{GM}{(R + h)^2} \\$We can solve these equations simultaneously. First, let's express $GM$ from the potential equation:$\\ GM = -V(R + h) \\$Substitute $GM$ in the equation for $g: \\ g = \frac{-V(R + h)}{(R + h)^2} \\$Simplify:$\\ g = \frac{-V}{R + h} \\$Rearrange to solve for $R + h: \\ R + h = \frac{-V}{g} \\$Substitute the given values:$\\ R + h = \frac{-(-5.4 \times 10^7)}{6.0} \\ R + h = \frac{5.4 \times 10^7}{6.0} \\ R + h = 0.9 \times 10^7 \, \mathrm{m} \\$Now, calculate $h: \\ h = (0.9 \times 10^7) - (6400 \times 10^3) \\ h = 9000 \times 10^3 - 6400 \times 10^3 \\ h = 2600 \times 10^3 \, \mathrm{m} \\ h = 2600 \, \mathrm{km} \\$Therefore, the height from the surface of the Earth is $2600 \, \mathrm{km}. \\$This corresponds to Option 3.