METHOD 1.
$\rightarrow$Total no. of attempts combination possible for a student (= no. of holes =m)
= For each question he can choose any 1 of the 4 options and there are 5 such questions
$ = 4*4*4*4*4$
$ = 1024.$
$\rightarrow$ We need to find no. of students (= no. of pigeons = n)
$\rightarrow$ It is given that at least $4$ answer sheets will be identical (= no. of pigeons that share the same hole should be at least 4.)
According to pigeon hole principle ,
$\left \lfloor \frac{(n-1)}{m} \right \rfloor+1=4$
$\Rightarrow$ $\left \lfloor \frac{(n-1)}{1024} \right \rfloor =3$
$\Rightarrow n-1 =3 *1024=3072$
$\Rightarrow n=3073$
METHOD 2.
In total there are $4*4*4*4*4=1024$ combination of options selection possible and a student will choose any one combination out of it.
Questions is asking how many students should be there so that atleast $4$ of them selects the same combination.
suppose $1024$ students comes and pick $1024$ unique combinations.
then another $1024$ students come and pick $1024$ unique combinations
then another $1024$ students come and pick $1024$ unique combinations.
So uptil here each combination is selected by $3$ students and total students are $1024+1024+1024 = 3072.$
Now if a student comes and select a combination i.e. select any one of $1024$ combinations then we will have $1$ combination that is selected by $4$ student.
so minimum students required $= 3072+1 = 3073$