Mean = Mode = Median = 3

Since the mode is $3$ we know that there is at least one occurrence of the value $3$ . Next, since there are 10 students, the median will be the average of the 5th and 6th values going from lowest to highest. This can occur if these values were $(2,4)$ or $(3,3)$ , but if it were the first of these then the possibility of a value of $3$ would be ruled out. Thus the middle two values are $3$ .

So let the ascending sequence of values be $2,a,b,c,3,3,d,e,f,7$ , (as we are given that values of $2$ and $7$ definitely occur). Then as the mean must also be $3$ we must have that

$\dfrac{2 + a + b + c + 3 + 3 + d + e + f + 7}{10} = 3 \Longrightarrow a + b + c + d + e + f = 15$ ,

where $2 \le a \le b \le c \le 3$ and $3 \le d \le e \le f \le 7$ . But as the minimum value we can (uniquely) achieve under these conditions is $2 + 2 + 2 + 3 + 3 + 3 = 15$ we must have $a = b = c = 2$ and $d = e = f = 3$ .

The desired concatenation is thus $\boxed{2222333337}$ .