6. The function $f(x)$ is an odd function on $\mathbf{R}$, and $g(x)$ is a periodic function on $\mathbf{R}$ with a period of 4. It is known that
$$
f(-2)=g(-2)=6 \text {, }
$$

and $\frac{f(f(2)+g(2))+g(f(-2)+g(-2))}{g^{2}(20 f(2))}=\frac{1}{2}$.
Then the value of $g(0)$ is ( ).
(A) 2
(B) 1
(C) 0
(D) -1
6. A.

From $f(x)$ being an odd function, we get
$$
f(2)=-f(-2)=-6, f(0)=0 .
$$

From $g(x)$ being a periodic function with a period of 4, we get
$$
\begin{array}{l}
g(2)=g(-2)=6 . \\
\text { Also, } g(20 f(2))=g(-120) \\
=g(4 \times(-30))=g(0),
\end{array}
$$

Then, from the given equation, we have
$$
\frac{f(-6+6)+g(6+6)}{g^{2}(0)}=\frac{1}{2} .
$$

Thus, $\frac{f(0)+g(3 \times 4)}{g^{2}(0)}=\frac{1}{2}$, which gives
$$
\frac{g(0)}{g^{2}(0)}=\frac{1}{2} \Rightarrow g^{2}(0)=2 g(0) \text {. }
$$

Solving this, we get $g(0)=0$ (discard) or 2.