The possible value of $x$ satisfying the equation $\frac{(143÷13\times x-3\times 3)}{(6^2-7\times 5+x^2)}=1$ is:

Simplify the numerator and the denominator of the given expression:
Numerator:

$$143÷13\times x-3\times 3=11\times x-9=11x-9$$

Denominator:

$$6^2-7\times 5+x^2=36-35+x^2=1+x^2$$

Equate the simplified fraction to 1:

$$\frac{11x-9}{x^2+1}=1$$

$$11x-9=x^2+1$$

Rearrange into a standard quadratic equation form ($a{x}^2+bx+c=0$):

$$x^2-11x+10=0$$

Factorize the quadratic equation:

$$x^2-10x-x+10=0$$

$$x(x-10)-1(x-10)=0$$

$$(x-1)(x-10)=0$$

Thus, the possible values of $x$ are $x=1$ or $x=10$. Looking at the options, $1$ is the correct satisfying value.