Let the Event that ball happen to be black and green=E
Let the Event that ball taken from $\text{URN I}=U_I$
Let the Event that ball taken from $\text{URN II}=U_{II}$
Let the Event that ball taken from $\text{URN III}=U_{III}$
$$P(\frac{U_I}{E})=\frac{P(\frac{E}{U_I})\times P(U_I)}{P(\frac{E}{U_I})\times P({U_{I}})+P(\frac{E}{U_{II}})\times P({U_{II}})+P(\frac{E}{U_{III}})\times P({U_{III}})}$$
$$=\frac{\frac{\binom{2}{1}\times \binom{2}{1} }{\binom{7}{2}}\times \frac{1}{3}}{\frac{\binom{2}{1}\times \binom{2}{1} }{\binom{7}{2}}\times \frac{1}{3}+\frac{\binom{3}{1}\times \binom{4}{1} }{\binom{9}{2}}\times \frac{1}{3}+\frac{\binom{2}{1}\times \binom{2}{1} }{\binom{9}{2}}\times \frac{1}{3}}=\frac{3}{10}$$
$$P(\frac{U_{II}}{E})=\frac{P(\frac{E}{U_I})\times P(U_{II})}{P(\frac{E}{U_I})\times P({U_{I}})+P(\frac{E}{U_{II}})\times P({U_{II}})+P(\frac{E}{U_{III}})\times P({U_{III}})}$$
$$=\frac{\frac{\binom{3}{1}\times \binom{4}{1} }{\binom{9}{2}}\times \frac{1}{3}}{\frac{\binom{2}{1}\times \binom{2}{1} }{\binom{7}{2}}\times \frac{1}{3}+\frac{\binom{3}{1}\times \binom{4}{1} }{\binom{9}{2}}\times \frac{1}{3}+\frac{\binom{2}{1}\times \binom{2}{1} }{\binom{9}{2}}\times \frac{1}{3}}=\frac{21}{40}$$
$$P(\frac{U_{II}}{E})=\frac{P(\frac{E}{U_II})\times P(U_{II})}{P(\frac{E}{U_I})\times P({U_{I}})+P(\frac{E}{U_{II}})\times P({U_{II}})+P(\frac{E}{U_{III}})\times P({U_{III}})}$$
$$=\frac{\frac{\binom{3}{1}\times \binom{4}{1} }{\binom{9}{2}}\times \frac{1}{3}}{\frac{\binom{2}{1}\times \binom{2}{1} }{\binom{7}{2}}\times \frac{1}{3}+\frac{\binom{3}{1}\times \binom{4}{1} }{\binom{9}{2}}\times \frac{1}{3}+\frac{\binom{2}{1}\times \binom{2}{1} }{\binom{9}{2}}\times \frac{1}{3}}=\frac{21}{40}$$
$$P(\frac{U_{III}}{E})=\frac{P(\frac{E}{U_III})\times P(U_{III})}{P(\frac{E}{U_I})\times P({U_{I}})+P(\frac{E}{U_{II}})\times P({U_{II}})+P(\frac{E}{U_{III}})\times P({U_{III}})}$$
$$=\frac{\frac{\binom{2}{1}\times \binom{2}{1} }{\binom{9}{2}}\times \frac{1}{3}}{\frac{\binom{2}{1}\times \binom{2}{1} }{\binom{7}{2}}\times \frac{1}{3}+\frac{\binom{3}{1}\times \binom{4}{1} }{\binom{9}{2}}\times \frac{1}{3}+\frac{\binom{2}{1}\times \binom{2}{1} }{\binom{9}{2}}\times \frac{1}{3}}=\frac{9}{40}$$