$4$ balls can be chosen from $10$ balls in $^{10}C_4$ ways. (Total number of ways)
The desirable(favourable) case is choosing $2$ red balls from $6$ red balls and choosing $2$ green balls from $4$ green balls.
So the required probability would be: $\frac{^{6}C_2\times ^{4}C_2}{^{10}C_4}$ $=$ $\frac{3}{7}$
Option A is correct.