We are given: $|\vec a|=1,\quad |\vec b|=4,\quad \vec a\cdot \vec b=2$ and $\vec c=(2\,\vec a\times \vec b)-3\vec b$

We need to find: $\vec b\cdot \vec c$

Substitute the value of $\vec c$: $\vec b\cdot \vec c=\vec b\cdot \left((2\vec a\times \vec b)-3\vec b\right)$

Use distributive property of dot product: $\vec b\cdot \vec c=2\,\vec b\cdot (\vec a\times \vec b)-3(\vec b\cdot \vec b)$

Now, $\vec b\cdot (\vec a\times \vec b)=0$ because $\vec a\times \vec b$ is perpendicular to $\vec b$.

Also, $\vec b\cdot \vec b=|\vec b|^2=4^2=16$

Therefore, $\vec b\cdot \vec c=2(0)-3(16)=-48$

Hence the correct option is: $\boxed{-48}$ which is Option A.