jmc

Because $△JKL$ is a right triangle, $\tan K=\frac{JL}{JK}$. So $\tan K=\frac{3}{2}=\frac{JL}{2}$. Then $JL=3$.

By the Pythagorean Theorem, $KL=\sqrt{J{L}^2+J{K}^2}=\sqrt{3^2+2^2}=\boxed{\sqrt{13}}$.