On cyclic right $R/I$-modules for two sided ideal $I$ of $R$ Let $R$ be a ring such that every cyclic right $R$ module is either injective or projective and let $I$ be a two sided ideal of $R$ . Then why is every cyclic right $R/I$-module either injective  or projective ? And if $R/I$ is injective , then why is every cyclic right $R/I$-module injective ? This is Lemma 1 in http://www.ams.org/journals/proc/1975-053-01/S0002-9939-1975-0382349-X/S0002-9939-1975-0382349-X.pdf
, and there it is said that the fact is clear from the paper http://www.ams.org/journals/proc/1968-019-06/S0002-9939-1968-0231857-7/S0002-9939-1968-0231857-7.pdf
; but I couldn't find anything in that paper to deduce this facts . Please help . Thanks in advance 

A cyclic $R/I$-module is a cyclic $R$-module on which $I$ acts as zero. If $M$ is an $R/I$-module that is injective (or projective) as an $R$-module then it is injective (or projective) as an $R/I$-module, since every injective map $M\to X$ of $R/I$-modules is an injective map of $R$-modules, and therefore splits.
If $R/I$ is injective as an $R/I$-module and $P$ is a cyclic $R/I$-module that is projective, then the natural surjection $R/I\to P$ splits, and so $P$ is also an injective $R/I$-module.
The reference to Osofsky's paper is not relevant for this, but for the second part of Lemma 1, that you didn't reproduce.
