Given situation is shown in the figure. Let acceleration due to gravity at the surface of the planet be g. At height h above planet's surface v = 0.

According to question, acceleration due to gravity of the planet at height h above its surface becomes g/4.

$${g_h} = {g \over 4} = {g \over {{{\left( {1 + {h \over R}} \right)}^2}}}$$

$$4 = {\left( {1 + {h \over R}} \right)^2} \Rightarrow 1 + {h \over R} = 2$$

$${h \over R} = 1 \Rightarrow h = R$$.

So, velocity of the bullet becomes zero at h = R.

Also, $${v_{esc}} = v\sqrt N \Rightarrow \sqrt {{{2GM} \over R}} = v\sqrt N $$ ...... (i)

Applying energy conservation principle,

Energy of bullet at surface of earth = Energy of bullet at highest point

$${{ - GMm} \over R} + {1 \over 2}m{v^2} = {{ - GMm} \over {2R}}$$

$${1 \over 2}m{v^2} = {{GMm} \over {2R}}$$ $$\therefore$$ $$v = \sqrt {{{GM} \over R}} $$

Putting this value in eqn. (i), we get

$$\sqrt {{{2GM} \over R}} = \sqrt {{{NGM} \over R}} $$

$$\therefore$$ N = 2