Let $B$ measurable. I know that $$\mathbb E[X\mid B]:=\frac{\mathbb E[X\boldsymbol 1_B]}{\mathbb P(B)}.$$
So, I guess that $$\mathbb E[X\mid \boldsymbol 1_B]=\mathbb E[X\mid B]\boldsymbol 1_B+\mathbb E[X\mid B^c]\boldsymbol 1_{B^c},$$ but I'm not so sure. My idea is the following : $$\mathbb E[X\mid \boldsymbol 1_B=1]=\mathbb E[X\mid B]\quad \text{and}\quad \mathbb E[X\mid \boldsymbol 1_{B^c}]=\mathbb E[X\mid B^c].$$
But in the other side, if this would be true, we would have $\mathbb E[X\mid \boldsymbol 1_B]=\mathbb E[X\mid \boldsymbol 1_{B^c}],$ and I really don't think it's true. So, any idea about $\mathbb E[X\mid \boldsymbol 1_B]$ ?
Why do you think $\mathbb E[X\mid \boldsymbol 1_B]=\mathbb E[X\mid \boldsymbol 1_{B^c}]$ is wrong ? In fact, you have a more general statement that is $$\mathbb E[X\mid Y]:=\mathbb E[X\mid \sigma (Y)]\tag{1}.$$
If you are not familiar with this notation, it's not a problem. But what $(1)$ says is namely that the conditional expectation $\mathbb E[X\mid Y]$ depend rather $\sigma (Y)$ (the smallest $\sigma -$algebra that make $Y$ measurable) than on the r.v. $Y$. This mean that even if $Y(\omega )\neq Z(\omega )$ for all $\omega \in \Omega $, as far as $\sigma (Y)=\sigma (Z)$, then $\mathbb E[X\mid Y]$ and $\mathbb E[X\mid Z]$ are the same. Since $$\sigma (\boldsymbol 1_B)=\{\Omega ,B,B^c,\varnothing \}=\sigma (\boldsymbol 1_{B^c}),$$ the fact that $$\mathbb E[X\mid \boldsymbol 1_B]=\mathbb E[X\mid \boldsymbol 1_{B^c}],$$ should not be surprinsing.