Consider the situ

Mass of the block on the table, m₁ = 4 kg

Mass of the block hanging from the table, m₂ = 1 kg

Acceleration of the system = a

Initially when the system was at rest, speed u = 0 m/s

Speed at a later time, v = 0.3 m/s

Distance travelled, s = 1 m

From equation of motion, 2as = v²-u²

⇒ 2×a×1 = 0.3×0.3 - 0

⇒ a = 0.045 m/s²

For the 1 kg block hanging from the table, Let the tension in the string be 2T

m₂a = m₂g – 2T

2T = 1×(0.045 + 9.8)

2T = 9.845

T = 4.9225 N

For the 4kg block on the table, Let the force of friction be F and tension in the string will be T

T – F = 2m₁a

If coefficient of friction is μ, F = μm₁g

T – μm₁g = 2m₁a

4×μ×9.8 = 4.9225 - 2×4×0.045

39.2μ = 4.5625

μ = 0.116 ≈ 0.12

∴ Coefficient of friction between the block and the table = 0.12