consider motion of two balls with respect to rocket

Maximum distance of ball A from left wall $=\frac{ u ^2}{2 a }=\frac{0.3 \times 0.3}{2 \times 2}=\frac{0.09}{4} \approx 0.02 m$ so collision of two balls will take place very near to left wall

For $B \quad S = ut +\frac{1}{2} a t^2$

$\quad-4=-0.2 t-\left(\frac{1}{2}\right) 2 t^2 \quad \Rightarrow \quad t^2+0.2 t-4=0 $

$\Rightarrow \quad t=\frac{-0.2 \pm \sqrt{0.04+16}}{2}=1.9 $

$\text { nearest integer }=2 s$