A right $R$-module $B$ is flat if and only if $0\to B\otimes_R I\to B\otimes_R R$ is exact for every finitely generated left ideal $I$.
The proof of this statement is given in Rotman Advanced Modern Algebra Part 1. My question is on if direction so I only state if part of the proof: Conversely, Proposition B-4.103 shows that every (not necessarily finitely generated) left ideal $I$ is flat (for every finitely generated ideal contained in $I$ is flat). There is an exact sequence $Hom(B\otimes_RR,\Bbb Q/\Bbb Z)\to Hom(B\otimes_RI,\Bbb Q/\Bbb Z)\to 0$ that, by the Adjoint isomorphism, gives exactness of $Hom_R(R,Hom(B,\Bbb Q/\Bbb Z))\to Hom_R(I,Hom(B,\Bbb Q/\Bbb Z))\to 0$. This says that every map from a left ideal $I$ to $Hom(B,\Bbb Q/\Bbb Z)$ extends to a map $R\to Hom(B,\Bbb Q/\Bbb Z)$; thus, $Hom(B,\Bbb Q/\Bbb Z)$ satisfies the Baer criterion so $Hom(B,\Bbb Q/\Bbb Z)$ is injective. By Proposition B-4.108, $B$ is flat.
I think the first sentence of the proof implies $0\to B\otimes_RI\to B\otimes_RR$ is exact for every left ideal $I$ but I don't know why. Proposition B-4.103 states that If every finitely generated submodule of a right $R$-module $M$ is flat, then $M$ is flat. But the given assumption does not guarantee that finitely generated left ideal $I$ is flat.