Answer
The IUPAC name of the alkene is $trans-3,4-dimethylcyclopent-1-ene$
Work Step by Step
The compound has 5 carbon atoms in its longest carbon ring containing the double bond ($pent-$). It has 2 $methyl-$ substituents at positions 3 & 4, and a double bond at position 1. The methyl groups are on opposite sides of the plane, so the compound is $trans$. So, the IUPAC name of the alkene is $trans-3,4-dimethylcyclopent-1-ene$