Since students watch either 1 or all 3:
let a = students who watched only A,
b = students who watched only B,
c = students who watched only C,
n = students who watched A, B and C.
Given data implies
a + n = 16 ……….[1]
b + n = 13 ……….[2]
c + n = 19 ……….[3]
and since total strength of class is 40,
a+b+c+n = 40 ……….[4]
solving for n ( [1] + [2] + [3] – [4] )
2n = 8
=> n=4