My attempt:
Consider the complement where all the consonants are considered as one unit: $BBBLLCS$
Step 1: Arrange the consonants within the unit: $\displaystyle\frac{7!}{3!2!}$
Step 2: Arrange the vowels: $\displaystyle\frac{6!}{2!2!}$
The total number of arrangements of the $13$ letters with no restriction is $\displaystyle\frac{13!}{3!(2!)^3}$
So the total number of arrangments where no two consonants are next to each other is
$$\displaystyle\frac{13!}{3!(2!)^3}-\left[\displaystyle\frac{7!}{3!2!}\cdot \displaystyle\frac{6!}{2!2!}\right]$$
There are 13 letters and 7 consonants. Since no two consonants can be together, this means there is a "space" between each consonant, dividing the possibilities into $C_ C_ C_ C_ C_ C_ C$ where C means Consonant. There is only one way for the general structure, but how do we choose which consonants for each spot? Since there are 3 B's, 2 L's, and one each of C and S, what does that give you? Hint: $\binom{7}{3}$ ways to place the B's, leaving 4 consonants; $\binom{4}{2}$ ways to place the L's, and so on.
I agree with you on arranging the vowels; since there is no other restriction, there are 6! ways to place vowels, but this double counts the U's (divide by $2$) and double counts the I's (divide by another 2).
The total is the product of the two.