Does there exist a concrete $C^*$ algebra $A$ such that that the following conditions hold:
(1) $A$ is unital and $A$ has no tracial state. (2)there exists a closed ideal $I$ of $A$ such that $I$ admits a tracial state and the center $Z(I)$ of $I$ is 0.
No, this does not exist, unless you allow the trivial solution $I = \{0\}$. Otherwise let $\tau$ be a tracial state on $I$ and let $(e_\lambda)$ be a quasi-central approximate unit for $I$. This means that $e_\lambda x - xe_\lambda \to 0$ for all $x \in A$. Then $(e_\lambda^{1/2})$ is also an approximate unit and $\phi: x \mapsto \lim \tau(e_\lambda x)$ is a tracial state on $A$.
The verification is straightforward. If $x \geq 0$ then $e_\lambda x - x^{1/2}e_\lambda x^{1/2} \to 0$, and $x^{1/2}e_\lambda x^{1/2}$ is increasing so $\tau(x^{1/2}e_\lambda x^{1/2})$ converges. This shows that $\tau(e_\lambda x)$ converges. Positivity of $\phi$ also follows because $\tau(x^{1/2}e_\lambda x^{1/2}) \geq 0$, and $\phi$ is a trace because, for any $x,y \in A$ we have $\tau(e_\lambda xy) \approx \tau(e_\lambda^{1/2} xe_\lambda^{1/2} y) = \tau(e_\lambda^{1/2} ye_\lambda^{1/2} x) \approx \tau(e_\lambda yx)$, where $\approx$ means the difference goes to zero in norm.
