Out of the three $2$ weight edges we have to take atleast $2$. Now, $ABE$ are connected.

Now if we select $BC=3$, then we have only $2$ options either $ED$ or $DC$.

If we don't select $BC$ then we can take any $2$ of $ED,EC\,or\,DC$

So total number of spanning trees $={}^3C_{2}*(2 + {}^3C_2)=15$