 Solution

If $x\ne b$, then $d_a(b,x)=d(b,a)+d(a,x)\ge d(b,a)$. The strict open-ball inequality therefore excludes every such $x$, while including $b$. Hence $\boxed{B_{d_a}(b,d(b,a))=\{b\}}$. Every point other than $a$ is isolated in this metric.