Solution:

To find the time period of a satellite at a distance \( 4R \) from the center of the Earth, we can use Kepler's third law of planetary motion, which states:

\[
T^2 \propto r^3
\]

1. Given:
- Time period at Earth's surface (\( T_0 \)): \( 1.4 \) hours (or \( T_0 = 1.4 \times 3600 \) seconds).
- Radius of the Earth: \( R \).

2. At Distance \( 4R \):
- The new radius \( r = 4R \).

3. Using the relationship:
\[
\frac{T^2}{T_0^2} = \frac{r^3}{R^3}
\]

Substituting \( r = 4R \):
\[
\frac{T^2}{T_0^2} = \frac{(4R)^3}{R^3} = \frac{64R^3}{R^3} = 64
\]

4. Solving for \( T \):
\[
T^2 = 64 T_0^2 ; T = 8 T_0
\]

5. Substituting \( T_0 \):
\[
T = 8\sqrt{2}, \text{hours} \]