Assume a ring $R$ is injective as an $R$-module. If the projective dimension of an $R$-module $P$ is finite could one conclude that $P$ is a projective $R$-module?
Probably one should start with a finite projective resolution for $P$, and then ...? Every free $R$-module is a direct sum of $R$ a cardinality many times. Now, how I can use the injectivity of $R$?
Let $R=K^{\mathbb N}$ be a countable direct product of copies of a field $K$. (For simplicity we may suppose $K=\mathbb Z/2\mathbb Z$.) This ring is self-injective, and its global dimension is $2$. But $I=K^{(\mathbb N)}$, a countable direct sum of copies of $K$, is an ideal of $R$ which is not a direct summand of $R$, and therefore $R/I$ isn't projective.
Edit. For those who don't like the continuum hypothesis and stuff, let me mention that isn't necessary to assume that $\operatorname{gldim}R=2$. The ideal $I$ is countably generated in a von Neumann regular ring, so it is projective (Kaplansky). This shows that $\operatorname{pd}_RR/I=1$.