Mass of monomer needed to obtain a polymer of a given molecular weight
You have a $2:1$ ratio. You need a $1:1$.

The $\bar{X}_m$ resulting (the number average degree of polymerisation) is derived from the Carothers equation:

$$\bar{X}_m = \frac{1+p}{1-p}$$

where $p$ is the mole ratio. For a $2:1$ this is $0.5$, and the effective $\bar{X}_m = 3$. If you mix at $2:1$, you'll have the $\ce{BAB}$ trimer as the major product with a $M_m$ of $\pu{297.1 g/mol}$.

If you have one component in $1\%$ excess ($1.01:1$) the $\bar{X}_m$ will be $199$, at $.1\%$ excess the $\bar{X}_m$ will be $1,999$ etc.