Yes. One can take $B$ to be the direct sum of all separable C*-algebras. A more interesting answer would be $B={\mathcal Q}(\ell_2)\otimes C^{\ast}(F_\infty)$. For an explanation, let me start with another question of you: It is an open problem
whether $B = {\mathcal Q}(\ell_2)$, the Calkin algebra, suffices or not.
It's written in my textbook with Nate Brown (Problem 10.4.1). If $A\odot{\mathcal Q}(\ell_2)$ has a unique C*-norm, then $A$ is exact and $A\odot{\mathcal B}(\ell_2)$ has also a unique C*-norm. The latter condition is equivalent to that $A$ has the LLP. Kirchberg's QWEP conjecture (which is equivalent to Connes's Embedding Conjecture) asserts that LLP implies WEP. It is known that exact (or local reflexivity) $+$ WEP implies nuclearity. In conclusion, if one finds a non-nuclear $A$ which has a unique C*-norm on $A \odot{\mathcal Q}(\ell_2)$, then one solves the QWEP conjecture in negative. Since WEP is equivalent to that $A \odot C^{\ast}(F_\infty)$ has a unique C*-norm, $B={\mathcal Q}(\ell_2)\otimes C^{\ast}(F_\infty)$, or any C*-algebra $B$ containing both ${\mathcal Q}(\ell_2)$ and $C^{\ast}(F_\infty)$ with conditional expectations, meets the condition. Whether there exists a separable $B$ that meets the condition is not known.