smc

If the areas of the two triangles are $A_1$ and $A_2$ with $A_1>A_2$, we are given that $A_1-A_2=18$ and $\frac{A_1}{A_2}=k^2$ Plugging in the second equation into the first leads to $(k^2-1)A_2=18$. If $A_2$ is an integer, it can only be a factor of $18$ - namely $1,2,3,6,9,18$. Since $A_1=A_2+18$, this would lead to $A_1=19,20,21,24,27,36$. This would lead to $\frac{A_1}{A_2}=19,10,7,4,3,2$, and the only square is $4$. Thus, the ratio of the areas is $4$, and hence the ratio of the sides is $\sqrt{4}=2$. The corresponding side has a length of $3\cdot 2=6$, which is option $\boxed{6}$.