A bullet is fired vertically upwards with velocity υ from the surface of a spherical planet. When it reaches its maximum height, its acceleration due to the planet's gravity is 1/4th of its value at the surface of the planet. If the escape velocity from the planet is υesc = υ√N, then the value of N is (ignore energy loss due to atmosphere)

We know that at the surface of the earth, value of g = GM/R²
At height h above the Earth's surface, the value of acceleration due to gravity g′ = GM/(R+h)²
So it is given that g′ = g/4
When the bullet reaches maximum height, acceleration due to gravity is 1/4th of that at planet's surface.
That implies GM/(4R²) = GM/(R+h)² → h = R
By conservation of mechanical energy,
-GMm/R + 1/2mv² = -GMm/(h+R) + 0
since velocity is zero at max height.
→ 1/2mv² = GMm/2R
v = √(GM/R) = √(2GM/2R) = 1/√2 √(2GM/R) = 1/√2 vesc
→ vesc = √2v
→ N = 2