Right projective but not bimodule projective

Question

Let $R$ be a ring with $1$. What is an example of a right $R$-module $M$ (and a ring $R$) such that $M$ is projective as a right $R$-module but not projective as an $R$-bimodule (assuming $M$ has structure of $R$-bimodule).

Answer 1

Take the regular bimodule $_RR_R$: it is free (hence projective) as a left/right $R$-module, but projective as an $R$-$R$-bimodule only if $\text{Hom}_{R-R}(_RR_R,-)\cong \text{HH}^0(R;-): M\mapsto \{m\in M\ |\ \forall r\in R: rm=mr\}$ is exact.

Example where this is not the case: Consider $R = {\mathbb Z}[X]$. Identifying $R$-$R$-bimodules with ${\mathbb Z}[X,Y]$-modules, the quotient map ${\mathbb Z}[X,Y]\to{\mathbb Z}[X,Y]/(X-Y)$ is surjective, but the symmetric element $\overline{1}\in{\mathbb Z}[X,Y]/(X-Y)$ does not lift to one in ${\mathbb Z}[X,Y]$.