We need to find |(vec b \times \vec a)-\vec b|^2. Given: $|\vec b|=1, \qquad |\vec b\times \vec a|=2.$

Use the identity $|\vec u-\vec v|^2=|\vec u|^2+|\vec v|^2-2\vec u\cdot \vec v.$ Let $\vec u=\vec b\times \vec a, \qquad \vec v=\vec b.$ Then |(vec b\times \vec a)-\vec b|^2=|\vec b\times \vec a|^2+|\vec b|^2-2(\vec b\times \vec a)\cdot \vec b.

Now, $\vec b\times \vec a$ is perpendicular to both $\vec b$ and $\vec a$. Hence $(\vec b\times \vec a)\cdot \vec b=0.$ So the expression becomes |(vec b\times \vec a)-\vec b|^2=|\vec b\times \vec a|^2+|\vec b|^2.

Substitute the given values: $|\vec b\times \vec a|^2=2^2=4, \qquad |\vec b|^2=1^2=1.$ Therefore, |(vec b\times \vec a)-\vec b|^2=4+1=5.

Hence the correct option is $\boxed{5}$ which is Option C.