jmc

We see that $(7b^3{)}^2=7^2\cdot b^{3\cdot 2}=49\cdot b^6.$ Likewise, $(4b^2{)}^{-3}=4^{-3}\cdot b^{-6}.$ Now, $(7b^3{)}^2\cdot (4b^2{)}^{-3}=49\cdot b^6\cdot 4^{-3}\cdot b^{-6},$ and since $4^{-3}=\frac{1}{64},$ we have $\frac{49}{64}\cdot b^6\cdot b^{-6}=\boxed{\frac{49}{64}},$ since $b^0=1$ for all non-zero $b.$