Given that in $\triangle ABC$, $a=2 \sqrt {3}$, $b= \sqrt {6}$, $A=45^{\circ}$,

By the sine law, we have: $\sin B= \dfrac {b\cdot \sin A}{a}= \dfrac { \sqrt {6}\times \dfrac { \sqrt {2}}{2}}{2 \sqrt {3}}= \dfrac {1}{2}$,

Since $b < a$, we can deduce that $B$ is an acute angle,

Hence, $B=30^{\circ}$.

Therefore, the answer is: $\boxed{30^{\circ}}$.

From the known information and the sine law, we can derive $\sin B= \dfrac {1}{2}$. By using the principle that the larger side is opposite the larger angle, we can determine that $B$ is an acute angle. Finally, by using the special angle's trigonometric function values, we can find the value of $B$.

This problem primarily tests the application of the sine law, the principle of the larger side being opposite the larger angle, and the trigonometric function values of special angles in solving triangles. It requires the ability to transform thinking and is considered a basic problem.