Calculate Huffman code length having probability?
The intended answer is probably that the length of the longest codeword is approximately $$-\log_2 10^{-6} = 20.$$ But this is wrong. The information given doesn't come close to specifying the length of the longest codeword. Even the entire probability distribution doesn't specify the length of the longest codeword. 

One can see this by constructing a Huffman tree with a probability distribution with probabilities proportional to the Fibonacci numbers $$\{1,1,1,2,3,5,8,13, \ldots, F_n\}.$$ (The third 1 is deliberate.) When combining these into a Huffman tree, you are faced with lots of choices. One set of choices yields a tree of depth approximately $n$; another, approximately $n/2$. 

For the probabilities $$\frac{1}{34}, \frac{1}{34}, \frac{1}{34}, \frac{2}{34}, \frac{3}{34}, \frac{5}{34}, \frac{8}{34}, \frac{13}{34},$$
we have the large-depth Huffman tree where the longest codeword has length 7:



and the small-depth Huffman tree where the longest codeword has length 4:



Both of these trees have $43/17$ for the expected length of a codeword, which is optimal.

It is possible to show that the length of the longest codeword is no more than a factor of $\log_\phi(2)\approx 1.44$ larger than $-\log_2 p$, where $\phi$ is the golden ratio; the worst case example is the large-depth one above, using Fibonacci numbers. So for your example the length is at most $28$. And because there are 1024 codewords, the length of the longest one has to be at least $\log_2 1024 = 10$.  So the correct answer is the length of the longest codeword is between $10$ and $28$. But unless the OP changed the wording of the question, these bounds don't seem to be what the problem was asking for.