2. Given real numbers $a, b, c$ satisfy the equations
$$
2 \sqrt{a}+3|b|=6, \quad 4 \sqrt{a}-9|b|=6 c \text {. }
$$

Then the maximum possible value of $c$ is ( ).
(A) 0
(B) 1
(C) 2
(D) 3
2. C.

From the two known equations, we get
$$
\sqrt{a}=\frac{3}{5}(c+3),|b|=\frac{2}{5}(2-c) \text {. }
$$

Since $|b| \geqslant 0$, therefore, $c \leqslant 2$.
When $c=2$, we get $a=9, b=0$, which satisfies the known equations.
Therefore, the maximum value that $c$ can take is 2.