To solve this problem, we need to analyze the enolic form of ethyl acetoacetate.$\\$The structure given in the image shows the enolic form on the left and the keto form on the right.$\\$Let's count the bonds in the enolic form:$\\$• $$Sigma bonds:$\\$ - Between each pair of atoms connected by a single line.$\\$ - Count the following:$\\$ - 3 C-H bonds in the $$CH$_3$ group.$\\$ - 1 C-H bond in the $$CH$$ group.$\\$ - 1 C-OH bond.$\\$ - 1 C-C bond between the $$CH$_3$ and $$CH$$ groups.$\\$ - 1 C-C bond between the $$CH$$ and $$C$$ groups.$\\$ - 1 C-C bond between the $$C$$ and $$CO$_2$H$_5$ groups.$\\$ - 1 C-O bond in the $$CO$_2$H$_5$ group.$\\$ - 2 C-H bonds in the $$CH$_2$ group of $$CO$_2$H$_5. \\$ - 1 C-C bond in the $$CO$_2$H$_5$ group.$\\$ - 1 C-O bond in the $$CO$_2$H$_5$ group.$\\$ - 1 O-H bond in the enol group.$\\$ - Total sigma bonds: 18$\\$• $$Pi bonds:$\\$ - 1 C=C bond in the enol form.$\\$ - 1 C=O bond in the $$CO$_2$H$_5$ group.$\\$ - Total pi bonds: 2$\\$Therefore, the enolic form of ethyl acetoacetate has 18 sigma bonds and 2 pi bonds.$\\$This corresponds to Option 4.