The more general phenomenon behind this is that $E$ is not a Cartier divisor at the ODPs. (A Cartier divisor could not be smooth, because that would imply that $X$ is smooth!) The pre-image of this divisor will be Cartier and hence the blow-up is not an isomorphism (as it is when one blows up a Cartier divisor). On the other hand the divisor is locally defined by $2$ equations near the ODPs, so the fibers over these points can be at most dimension $1$. Hence they are exactly dimension $1$. In other words, the blow up is a small morphism. Then you need a local computation to see that the blow up is indeed smooth. You may do this by verifying that the pre-image of $E$ is smooth which maybe(??) easier to do since $E$ is smooth to begin with.

The more general phenomenon behind this is that $E$ is not a Cartier divisor at the ODPs. (A Cartier divisor could not be smooth, because that would imply that $X$ is smooth!) The pre-image of this divisor will be Cartier and hence the blow-up is not an isomorphism (as it is when one blows up a Cartier divisor). On the other hand the divisor is locally defined by $2$ equations near the ODPs, so the fibers over these points can be at most dimension $1$. Hence they are exactly dimension $1$. In other words, the blow up is a small morphism. 

Then you need to see that the blow up is indeed smooth. You may do this by verifying that the pre-image of $E$ is smooth which should not be too hard as its pre-image ought to be just the blow up of E at the ODP points (check this by an explicit local computation). These points are smooth on $E$, so the blow-up of $E$ remains smooth. Then its a smooth Cartier divisor, so the ambient space has to be smooth along it, but it is already smooth everywhere else.