4. The largest integer not exceeding $(3 \sqrt{2}+2 \sqrt{3})^{6}$ is ( ).
(A) 209520
(B)209 519
(C)209 518
(D) 209517
4. B.

Let $3 \sqrt{2}+2 \sqrt{3}=a, 3 \sqrt{2}-2 \sqrt{3}=b$.
Then $a+b=6 \sqrt{2}, ab=6$
$$
\begin{array}{l}
\Rightarrow a^{2}+b^{2}=(a+b)^{2}-2ab=60 \\
\Rightarrow a^{6}+b^{6}=\left(a^{2}+b^{2}\right)^{3}-3(ab)^{2}\left(a^{2}+b^{2}\right) \\
\quad=209520 .
\end{array}
$$

Since $0<b<1$, we have
$$
209519<a^{6}<209520 \text {. }
$$

Therefore, the largest integer not exceeding $(3 \sqrt{2}+2 \sqrt{3})^{6}$ is
209519.