If $$\frac{1}{2+\frac{1}{3+\frac{1}{4+\frac{1}{\frac{1}{x}}}}}$$ = $$\frac{68}{157}$$, then the value of $$x^{3} + 3x^{2} - 9x + 5$$.

Solving the given relation, 
$$\frac{157}{68}=2+\frac{1}{3+\frac{1}{4+\frac{1}{x}}}$$
$$\frac{21}{68}=\frac{1}{3+\frac{1}{4+\frac{1}{x}}}$$
$$3+\frac{1}{4+\frac{1}{x}}=\frac{68}{21}$$
$$\frac{1}{4+\frac{1}{x}}=\frac{5}{21}$$
$$\frac{21}{5}=4+\frac{1}{x}$$
$$\frac{1}{5}=\frac{1}{x}$$

Giving the value of x to be 5, 
Using this we can find the value of polynomial to be 
$$5^3+3\left(5\right)^2-9\left(5\right)+5$$
$$125+75-45+5$$
= 160

Therefore, Option A is the correct answer.