Total students = 50

Students who play Hockey
a+d+g+e = 20 ---- (1)

Students who play Cricket
b+f+g+d = 15 ---(2)

Students who play Football
c+e+g+f = 11 ----- (3)

Both Hockey & Cricket
d+g = 7 ------ (4)

Both Cricket & Football
g+f = 4 ------(5)

Both Football & Hockey
g+e = 5 ------ (6)

18 students do not play any given sports,
then a+b+c+d+e+f+g = 50-18=32 -----(7)

We need to find number of students playing exactly two of these sports
i.e d+e+f

adding (1),(2) &(3)
a+b+c+2d+2e+2f+3g = 20+15+11=46 ------ (8)

subtracting (7) from (8)
d+e+f+2g=14 ---- (9)

adding (4),(5) &(6)
d+e+f+3g = 5+4+7 = 16 ----(10)

subtracting (9) from (10)
g=2

substituting value of g in (9)
d+e+f =10

hence ans is B

In problems such as these, using the equations is easier than using Venn-Diagrams.

Equation 1:
Total - neither = H + C + F - [H&C + C&F + F&H] + H&C&F
50 - 18 = 20 + 15 + 11 - [7 + 4 + 5] + H&C&F
32 = 30 + H&C&F
H&C&F = 2

Equation2:
Total - neither = H + C + F - [people in only two groups] - 2*H&C&F
(using the value of H&C&F from above)
50 - 18 = 20 + 15 + 11 - [people in exactly two groups] - 2*2
people in exactly two groups = 10

[spoiler](B)[/spoiler] is the answer.

Note:
One should have the three methods - (i) drawing Venn Diagrams, (ii) using the equations above, and (iii) the Table approach- in your repository so that you can choose one of the depending on what will be easier for the given problem.

I solved a similar problem a few weeks back here:
https://www.beatthegmat.com/the-fastest- ... tml#465320
where I also explained the derivation of the Equation 2.

Thanks Aneesh.