How much of ingredient $a$ ?

Fraction of ingredient $a$ in solution X is $\dfrac{2}{2+3} = \dfrac{2}{5}$

Fraction of ingredient $a$ in solution Y is $\dfrac{1}{1+2} = \dfrac{1}{3}$

Fraction of solution X in solution Z is $\dfrac{3}{3+11} = \dfrac{3}{14}$

Fraction of solution Y in solution Z is $\dfrac{11}{3 + 11} = \dfrac{11}{14}$

Therefore, amount of ingredient $a$ in $2520$ ounces of solution Z is:

$m = \displaystyle 2520 \cdot \left( \left( \dfrac{3}{14} \right) \left( \dfrac{2}{5} \right) + \left( \dfrac{11}{14} \right) \left( \dfrac{1}{3} \right) \right)$

$m = \displaystyle 2520 \cdot \left( \dfrac{3}{35} + \dfrac{11}{42} \right) = 2520 \cdot \left( \dfrac{3 \times 6 + 11 \times 5 }{210} \right)$

$m = 2520 \cdot \left( \dfrac{73}{210} \right) = 252 \left( \dfrac{73}{3 \times 7} \right) = 84 \times \dfrac{73}{7} = 12 \times 73 = 876$ ounces

Since solution $Z$ is created by mixing solutions $X$ and $Y$ in a ratio of $3:11$ , let solution $X$ have $3k$ ounces and let solution $Y$ have $11k$ ounces, so that $3k + 11k = 2520$ , which solves to $k = 180$ , so that there are $3k = 3 \cdot 180 = 540$ ounces of solution $X$ and $11k = 11 \cdot 180 = 1980$ ounces of solution $Y$ .

Since solution $X$ has ingredients $a$ and $b$ in a ratio of $2:3$ , let ingredient $a$ have $2k$ ounces and let ingredient $b$ have $3k$ ounces, so that $2k + 3k = 540$ , which solves to $k = 108$ , so there are $2k = 2 \cdot 108 = 216$ ounces of ingredient $a$ in solution $X$ in solution $Z$ .

Since solution $Y$ has ingredients $a$ and $b$ in a ratio of $1:2$ , let ingredient $a$ have $k$ ounces and let ingredient $b$ have $2k$ ounces, so that $k + 2k = 1980$ , which solves to $k = 660$ , so there are $k = 660$ ounces of ingredient $a$ in solution $Y$ in solution $Z$ .

Therefore, solution $Z$ has a total of $216 + 660 = \boxed{876}$ ounces of ingredient $a$ .

Solution X contains exactly 2 / (2 + 3) = 2/5 = 40% of ingredient A while solution Y contains around 1 / (1 + 2) = 1/3 ≈ 33% of ingredient A.
A new solution Z which is actually a mixture of both X and Y is created by the ratio of 3 : 11, so solution Y dominated by nearly 4 times the volume of solution X. This will make solution Z's percentage of ingredient A nearer to 33% than 40% by nearly 4 times to around 35%. This is still around one third of the total volume of Z, so 2520 / 3 ≈ 800+ to match the only possible option of 876.