smc

There are two cases: (i) $x-2>0$ and (ii) $x-2<0$. In case (i), $0<x-2<0.01$, so $4<x+2<4.01$. In case (ii), $-0.01<x-2<0$, so $3.99<x+2<4$. Because $(x+2)$ in case (i) can take a larger value, we use it to determine the upper bound for $|x^2-4|$ under the given restrictions. Now, we can see that $|x^2-4|=|x-2||x+2|<0.01\ast 4.01=\boxed{0.0401}$, which is answer choice $\boxed{.0401}$.