Solution:

The escape velocity \( V_e \) is the speed needed to escape Earth's gravitational pull. The energy conservation principle applies, where the total mechanical energy at the surface and at height \( h = 3R \) (where \( R \) is the Earth's radius) should be equal.

The total energy at the surface:
\[
E_1 = \frac{1}{2} m V_e^2 - \frac{G M m}{R}
\]

At height \( 3R \), the total energy is:
\[
E_2 = \frac{1}{2} m v^2 - \frac{G M m}{4R}
\]

Since \( E_1 = E_2 \), we equate the two:
\[
\frac{1}{2} m V_e^2 - \frac{G M m}{R} = \frac{1}{2} m v^2 - \frac{G M m}{4R}
\]

We know that the escape velocity is given by:
\[
V_e^2 = \frac{2 G M}{R}
\]

Substitute this into the equation:
\[
\frac{1}{2} m \frac{2 G M}{R} - \frac{G M m}{R} = \frac{1}{2} m v^2 - \frac{G M m}{4R}
\]

Simplifying:
\[
\frac{G M m}{R} - \frac{G M m}{R} = \frac{1}{2} m v^2 - \frac{G M m}{4R}
\]

This simplifies to:
\[
0 = \frac{1}{2} m v^2 - \frac{G M m}{4R}
\]

Solving for \( v^2 \):
\[
\frac{1}{2} m v^2 = \frac{G M m}{4R}
\]

\[
v^2 = \frac{2 G M}{4R} = \frac{G M}{2R}
\]

Thus, the velocity at height \( 3R \) is:
\[
v = \sqrt{\frac{G M}{2R}} = \frac{V_e}{\sqrt{2}}= 7.92 km/sec
\]

So, the velocity at height \( 3R \) is \( \frac{V_e}{\sqrt{2}} \).= 7.92 km/sec