Consider the motion of two balls with respect to rocket.

Distance travelled by ball A from left end of the chamber is

$$ = {{{u^2}} \over {2a}} = {{{{(0.3)}^2}} \over {2 \times 2}} = {{0.09} \over 4} \approx 0.02$$ m

So, collision of two balls will take place very near to left end of the chamber.

For ball B

$$S = ut{1 \over 2}a{t^2}$$

$$ - 4 = - 0.2 \times t - {1 \over 2} \times 2 \times {t^2}$$

$${t^2} + 0.2t - 4 = 0$$

$$t = {{ - 0.2 \pm \sqrt {{{(0.2)}^2} - 4(1)( - 4)} } \over 2} = {{ - 0.2 \pm \sqrt {0.04 + 16} } \over 2}$$

t = 1.9 s, $$-$$2.1 s

Since t can't be negative

$$\therefore$$ t = 1.9 s

Nearest integer is 2 s.

Also, from

$$S = ut + {1 \over 2}a{t^2}$$

$${S_A} = 0.3t + {1 \over 2}( - 2){t^2} = 0.3t - {t^2}$$

$${S_B} = 0.2t + {1 \over 2}(2){t^2} = 0.2t + {t^2}$$

$$\because$$ $${S_A} + {S_B} = 4$$

$$\Rightarrow$$ 0.5 t = 4 or t = 8 s