I'll give it a try-

From the given data, books read = A,C,D = 10+8+8 = 26
Remaining books to be read = 56-26= 30
Remaining students = 26-(10+8)=8

Now,if 7 students read 4 books (7*4=28), the remaining student can read the remaining (30-28)=2 books which is the answer.

[spoiler]Ans: 2[/spoiler]

There can be one doubt that what if 7 students read only one book while the remaining student reads 23 ?
Well, then it sounds illogical that this one students reads the same book several times.
Question should have been slightly more specific to include that 'a student can read one book only once'.

Hope this satisfies.

Difficult to draw a Venn diagram for this question, so let's try and do it more efficiently.

We have 26 students but we're asked about the remaining ones after subtracting 8 and 10 students -- hence 8 students left to consider.

Assuming each book is only read by one student, there are 30 books remaining from the 56 to be read by these 8 students (56 - 10 - 2x8).

There are 4 types of book, so each of these 8 students could read up to 4 books. To minimise one student's reading list, we maximise all the others and say that 7 students read 4 books each, leaving 2 for the last student. Answer: 2.

As adthedaddy points out, there are a few assumptions in the question that it would be better had been made explicit.

Ya, adthedaddy's solution is correct. I had a tough time in finding that assumption !