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Clearly it's a prime ideal containing the radical. Do you just call it the "Jacobson radical of I?"

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    I suppose in some sense it's the Jacobson radical of $R/I$.2012-11-27
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    Well, it's the *preimage* of the Jacobson radical of $R/I$ under the projection $R \to R/I$. Of course, we don't call the radical of an ideal the "nilradical of I," despite the analogous setup.2012-11-27
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    I guess if I wanted to be particularly obtuse, I could analogously drop the first three letters and call it the "obson radical," but that's just silly.2012-11-27
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    Yes, that's the sense I was alluding to. Although I might phrase it as the contraction of the Jacobson radical of R/I under the canonical projection. It seems though that Jacobson radical of an ideal, is a [defined notion](http://crazyproject.wordpress.com/2010/10/04/definition-and-basic-properties-of-the-jacobson-radical-of-an-ideal/).2012-11-27
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    The intersection you are interested in need **not** be a prime ideal. Counter-example: $I=(0)\subset R=\mathbb Z/(6)$. By the way, be very careful with the adverb *clearly* : it is a magnet for false statements.2013-05-18
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    Oops! I don't know how the word "prime" got there in the first place. (This question is several months old.) I must have been planning to write something like "clearly it's an ideal containing the intersection of the prime ideals containing I, i.e. the radical of I," and then screwed up the editing. Thanks for pointing it out, though -- it could definitely confuse someone who came along and read this.2013-05-18
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    By the way, I think it's easier to visualize the ideal corresponding to, say, a few points in the plane. Then obviously the Jacobson radical of that ideal still has to cut out just as many components and thus can't be prime.2013-05-18

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For the sake of resolving this question, it apparently is sometimes known as the "Jacobson radical of $I$." Thanks to JSchlather for supplying a reference in the comments.