You can't find the precise integer without knowing what the product is - your question is incomplete - but you can find a general solution.
If \(x\) is one of the integers, then the next even integer is given by \(x+2\). So if the product is \(n\), then you have
\(x(x+2)=k\implies x^2+2x=k\implies x^2+2x+1=k+1\implies (x+1)^2=k+1\implies x=-1\pm\sqrt{k+1}\)
Plug in \(k\) and select the solution for \(x\) that is an even integer.
