$i\sqrt{x+6\sqrt{2}}-\sqrt{x-6\sqrt{2}}=2\sqrt{2}$ , then $x$ equals

The correct option is 11.

Let's verify this step-by-step by substituting the correct value $x=11$ into the given equation:
$$\sqrt{x+6\sqrt{2}}-\sqrt{x-6\sqrt{2}}=2\sqrt{2}$$

First, we substitute $x=11$ into the expression inside the first square root:
$$11+6\sqrt{2}$$
We want to express this expression as a perfect square of the form ${(a+b\sqrt{2})}^2=a^2+2b^2+2ab\sqrt{2}$.
Let's rewrite the term $6\sqrt{2}$ as $2·3·\sqrt{2}$.
If we choose $a=3$ and $b=\sqrt{2}$, then:
$$a^2+b^2=3^2+{\left(\sqrt{2}\right)}^2=9+2=11$$
This perfectly matches our constant term $11$. Therefore, we can write:
$$11+6\sqrt{2}=3^2+{\left(\sqrt{2}\right)}^2+2·3·\sqrt{2}={(3+\sqrt{2})}^2$$

Similarly, for the second term, we substitute $x=11$:
$$11-6\sqrt{2}=3^2+{\left(\sqrt{2}\right)}^2-2·3·\sqrt{2}={(3-\sqrt{2})}^2$$

Now, taking the square roots of both expressions:
$$\sqrt{11+6\sqrt{2}}=\sqrt{{(3+\sqrt{2})}^2}=3+\sqrt{2}$$
And since $3>\sqrt{2}$:
$$\sqrt{11-6\sqrt{2}}=\sqrt{{(3-\sqrt{2})}^2}=3-\sqrt{2}$$

Finally, we subtract the two simplified terms:
$$(3+\sqrt{2})-(3-\sqrt{2})=3+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}$$
This matches the right-hand side of the original equation. Thus, the value $x=11$ satisfies the equation.