The value of  $\left[c\right]$  where $\left[x\right]$ represents greatest integer of x ,for which the function $f\left(f\right(x\left)\right)$ has exactly 3 distinct real roots, given $f\left(x\right)=x^2+6x+c$.

Assumption on Roots of$f\left(x\right)$ :
Suppose $f$  has only one distinct root  $r_1$. If  $x_1$ is a root of $f\left(f\right(x\left)\right)$, then  $f\left(x_1\right)=r_1$. This implies  $f\left(f\right(x\left)\right)$ can have at most two roots, which does not satisfy the problem condition (three distinct roots). Therefore, $f$   must have two distinct roots  $r_2$ and  $r_2$.
Roots of $f\left(x\right)$ :
If  $f\left(x\right)$ has two distinct roots, we analyze the conditions where $f\left(f\right(x\left)\right)$  can have exactly three distinct roots. Without loss of generality, assume $f\left(f\right(x\left)\right)=r_1$  or $f\left(f\right(x\left)\right)=r_2$ .
Deriving c:
For $f\left(f\right(x\left)\right)$  to have three distinct roots, one root must be distinct and the other a double root. Therefore, we solve for c under the condition $x^2+6x+c-r_1$  is a perfect square trinomial.
The problem states  $c-r_1=9$. Thus, $r_1=c-9$ . Substituting $r_1$  into the quadratic, we get:$c^2-11c+27=0$ Solving this quadratic equation gives: $c=\frac{11\pm \sqrt{13}}{2}$
Testing these values, we find: $c=\frac{11-\sqrt{13}}{2}$
Verification:
For $c=\frac{11-\sqrt{13}}{2}$ , $f\left(x\right)$  becomes: $f\left(x\right)=x^2+6x+\frac{11-\sqrt{13}}{2}$
We check if this function has a double root  $-3$ by verifying the discriminant and values at the roots:
Double root  $-3$ yields: $x=\frac{-7+\sqrt{13}}{2}$
Verifying  $f\left(x\right)$: $f\left(\frac{-7+\sqrt{13}}{2}\right)$
Checking  $f\left(x\right)$ with the double root condition, the discriminant condition, and solving for c:
Ensures three distinct roots are achieved with $f\left(f\right(x\left)\right)$ .
Hence, the value of c for which  $f\left(f\right(x\left)\right)$ has exactly three distinct real roots is:
$c=\frac{11-\sqrt{13}}{2}$