Step-by-step Solution:

1. Determine the maximum height reached by the bullet.

   The acceleration due to gravity ($g$) at a distance $r$ from the center of a planet of mass $M$ is given by $g(r) = {{GM} \over {{r^2}}}$.

   At the surface of the planet, the distance from the center is the radius $R$, so the acceleration due to gravity is $g_s = {{GM} \over {{R^2}}}$.

   Let the maximum height reached by the bullet be $h$. At this height, the distance from the center of the planet is $r = R + h$. The acceleration due to gravity at this height is $g' = {{GM} \over {{(R + h)^2}}}$.

   The problem states that at the maximum height, the acceleration due to gravity is ${\left( {{1 \over 4}} \right)^{th}}$ of its value at the surface. Therefore: $g' = {1 \over 4}g_s$ ${{GM} \over {{(R + h)^2}}} = {1 \over 4} \cdot {{GM} \over {{R^2}}}$

   Canceling $GM$ from both sides: ${1 \over {{(R + h)^2}}} = {1 \over {4{R^2}}}$

   Taking the reciprocal and then the square root of both sides: $(R + h)^2 = 4R^2$ $R + h = 2R$ $h = R$

   So, the maximum height reached by the bullet is equal to the radius of the planet.

2. Apply the principle of conservation of mechanical energy.

   Since we ignore energy loss due to the atmosphere, the total mechanical energy of the bullet is conserved. The total energy at the surface (initial state) must equal the total energy at the maximum height (final state).

   Initial State (at the surface):

   • Initial velocity = $v$
   • Kinetic Energy, ${K_i} = {1 \over 2}m{v^2}$ (where $m$ is the mass of the bullet)
   • Potential Energy, ${U_i} = - {{GMm} \over R}$
   • Total Initial Energy, ${E_i} = {K_i} + {U_i} = {1 \over 2}m{v^2} - {{GMm} \over R}$

   Final State (at maximum height, $h=R$):

   • At maximum height, the velocity is momentarily zero, so final velocity = 0.
   • Kinetic Energy, ${K_f} = 0$
   • Potential Energy, ${U_f} = - {{GMm} \over {R + h}} = - {{GMm} \over {R + R}} = - {{GMm} \over {2R}}$
   • Total Final Energy, ${E_f} = {K_f} + {U_f} = 0 - {{GMm} \over {2R}} = - {{GMm} \over {2R}}$

   By conservation of energy, ${E_i} = {E_f}$: ${1 \over 2}m{v^2} - {{GMm} \over R} = - {{GMm} \over {2R}}$

   Adding ${{GMm} \over R}$ to both sides: ${1 \over 2}m{v^2} = {{GMm} \over R} - {{GMm} \over {2R}} = {{GMm} \over {2R}}$

   Canceling $m$ and multiplying by 2: ${v^2} = {{GM} \over R}$

3. Relate the firing velocity to the escape velocity.

   The escape velocity from the surface of the planet is given by the formula: ${v_{esc}} = \sqrt {{{2GM} \over R}}$

   The problem states that ${v_{esc}} = v\sqrt N$. Squaring both sides gives: ${v_{esc}}^2 = {v^2}N$

   Now, substitute the expressions we found for ${v_{esc}}^2$ and ${v^2}$: ${{2GM} \over R} = \left( {{{GM} \over R}} \right)N$

   Dividing both sides by ${{GM} \over R}$: $N = 2$

   Therefore, the value of N is 2.