A box contains 8 red and $x$ number of green balls. 3 balls are drawn at random, if the probability that 3 balls being red is $\frac{7}{15}$ then number of green balls is _______

Let $R = 8$ be the number of red balls and $G = x$ be the number of green balls.
The total number of balls is then $N = R + G = 8 + x$.

The total number of ways to draw 3 balls is $\binom{8 + x}{3} = \frac{(8 + x)(7 + x)(6 + x)}{6}$.
The number of ways to draw exactly 3 red balls is $\binom{8}{3} = 56$.

Given that the probability of drawing 3 red balls is $\frac{7}{15}$, the equation becomes:
$\frac{7}{15} = \frac{56}{\frac{(8 + x)(7 + x)(6 + x)}{6}} = \frac{336}{(8 + x)(7 + x)(6 + x)}$.

Rearranging gives:
$(8 + x)(7 + x)(6 + x) = \frac{336 \times 15}{7} = 720$.

The product of three consecutive integers $(6 + x)$, $(7 + x)$, and $(8 + x)$ must equal 720.
Testing options:

• $x = 2$: $10 \cdot 9 \cdot 8 = 720$
• $x = 4$: $12 \cdot 11 \cdot 10 = 1320$
• $x = 3$: $11 \cdot 10 \cdot 9 = 990$
• $x = 5$: $13 \cdot 12 \cdot 11 = 1716$

Only $x = 2$ satisfies the condition, so the number of green balls is 2.
Final answer: $\boxed{\text{A}}$