The Boolean function $\text{F(A, B, C, D)}$ evaluates to $1$ for minterms $1$ and $9$, and it evaluates to $0$ for minterms $0$ and $12$.
Thus, Option D) $\mathrm{F}(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D})=\Sigma(1,3,4,6,7,8,9,11)$ is correct.