Are the two objects s acceleration same or different?

s = ut + (1/2)at²

For the faster object :
u = 0 m/s
a = a₁
Then, s₁ = 0 + (1/2)a₁t²
s₁ = (1/2)a₁t² …… [1]

For the slower object :
u = 0 m/s
a = a₂
Then, s₁ = 0 + (1/2)a₂t²
s₂ = (1/2)a₂t² …… [2]

[1] - [2] :
s₁ - s₂ = (1/2)(a₁ - a₂)t²

When t = 3 s, s₁ - s₂ = 12 m :
12 = (1/2)(a₁ - a₂)(3)²
a₁ - a₂ = 8/3
Then, s₁ - s₂ = (1/2)(8/3)t²
s₁ - s₂ = (4/3)t²

When t = 6 s :
Distance apart, s₁ - s₂ = (4/3)(6)² m = 48 m

Different, obviously.
In 3s. one has moved 12m. relative to the other. You could assume 1 does not move at all for the question, this is a 'relative' thing.
Acceleration = (2d/t^2) = (24/9) = 2.667m/sec^2.
After 6 secs. they will be 1/2 (t^2 x a) = 48 metres apart.

With constant acceleration and 0 starting velocity, distance is proportional to square of time s=0.5at^2.

6 s is twice as much as 3 s, so the distance covered by each object will be 4 times greater and the difference will be 12*4 = 48 m