Let number of red only balls be $a$ and green only ball be $b$ and both red and green be $c$.

So we can say $a+b+c=87$.

2/7 of the balls that have red color also have green color

So $\frac{c}{(a+c)}= \frac{2}{7}$.

3/7 of the balls that have green color also have red color

So $\frac{c}{(b+c)}=\frac{3}{7}$.

Inverting both fractions and adding them

$\frac{(a+c)}{c}+\frac{(b+c)}{c}=\frac{7}{3}+\frac{7}{2}$

Subtracting 1 from both sides

$\frac{a+b+2c}{c}-1= \frac{35}{6}-1$.

$\frac{a+b+c}{c}=\frac{29}{6}$

Inverting again

$\frac{c}{a+b+c}=\frac{6}{29}$.

Our objective was always to find $\frac{c}{a+b+c}$.

Hence option D is correct.