We have a bag containing $3$ red and $3$ green balls. Now, he draws three balls from the bag. So, possible cases for the event is as follows:

Clearly, from case $\left(3\right)$ and $\left(4\right)$, we cannot get balls of different colours. Hence, probability is zero.

Now,

probability of drawing $1$ red and $2$ green balls is

$=\frac{{}^{3}C_1·{}^{3}C_2}{{}^{6}C_3}=\frac{9}{20}$.

Then, $P\left(1\right)=\frac{{}^{2}C_1·{}^{1}C_1·{}^{3}C_1}{{}^{6}C_3}=\frac{6}{20}$.

And,

probability of drawing $2$ red and $1$ green ball is

$=\frac{{}^{3}C_1·{}^{3}C_2}{{}^{6}C_3}=\frac{9}{20}$.

Then, $P\left(2\right)=\frac{{}^{1}C_1·{}^{2}C_1·{}^{3}C_1}{{}^{6}C_3}=\frac{6}{20}$.

Hence, required potability is

$=\left(\frac{9}{20}\times \frac{6}{20}\right)+\left(\frac{9}{20}\times \frac{6}{20}\right)$.

$=2\left(\frac{9}{20}\times \frac{6}{20}\right)=\frac{27}{100}$ or $27\%$.