It is known $(r+1)^{\text {th }}$ term, $T_{r+1},$ in the binomial expansion of $(a+b)^{n}$ is  given by ${T_{r + 1}} = {\,^n}{C_r}{a^{n – r}}{b^r}$

Thus, the $4^{\text {th }}$ term in the expansion of $\left(x^{2}-2 y\right)^{12}$ is

${T_4} = {T_{3 + 1}} = {\,^{12}}{C_3}{(x)^{12 – 3}}{( – 2y)^3} = {( – 1)^3} \cdot \frac{{12!}}{{3!9!}} \cdot {x^9} \cdot {(2)^3} \cdot {y^3}$

$=-\frac{12 \cdot 11 \cdot 10}{3 \cdot 2} \cdot(2)^{3} x^{9} y^{3}=-1760 x^{9} y^{3}$