[tex] \large\begin{gathered} {\underline{\boxed{ \rm {\red{Volume \: of \: Cone \: = \: \pi \: {r}^{2} \: \frac{h}{3} }}}}}\end{gathered}[/tex]
[tex] \bf{\red{ \longrightarrow}} \tt \: r \: = \: \frac{Diameter}{2} \\ [/tex]
[tex]\bf{\red{ \longrightarrow}} \tt \: r \: = \: \frac{16 \: yd}{2} \\ [/tex]
[tex]\bf{\red{ \longrightarrow}} \tt \: r \: = \: \frac{ \cancel{16 \: yd} \: \: ^{8} }{ \cancel{2}} \\ [/tex]
[tex]\bf{\red{ \longrightarrow}} \tt \: \large{\bf{{{\color{navy}{r \: = \: 8 \: yd}}}}}[/tex]
[tex]\bf{\red{ \longrightarrow}} \tt \: \: \large{\bf{{{\color{navy}{h \: = \: 16 \: yd}}}}}[/tex]
[tex] \bf \large \longrightarrow \: \: 3.14 \: \times \: {8}^{2} \: \times \: \frac{16}{3} \\ [/tex]
[tex]\bf \large \longrightarrow \: \:3.14 \: \times \: 64 \: \times \: \frac{16}{3} \\ [/tex]
[tex]\bf \large \longrightarrow \: \:3.14 \: \times \: 64 \: \times \: \frac{ \cancel{16} \: \: ^{5.33} }{ \cancel{3}} \\ [/tex]
[tex]\bf \large \longrightarrow \: \:3.14 \: \times \: 64 \: \times \: 5.33[/tex]
[tex]\bf \large \longrightarrow \: \:200.96 \: \times \: 5.3[/tex]
[tex]\bf \large \longrightarrow \: \:1071.79[/tex]
Option (A) is the correct answer
