From the problem, we have 

    100*A + 10*B + C = 16*(10A + C),   (1)

    1 <= A <= 9,   0 <= B, C <= 9.     (2)


Equation (1) is equivalent to

    100*A + 10*B + C = 160*A + 16C,

    60*A  - 10B + 15C = 0,

    12*A - 2B + 3C = 0.      (3)


From the last equation (3), we conclude that 

    (a)  C is an even number 0, 2, 4, 6, or 8; 

    (b)  B is a multiple of 3, so B is either 0, or, 3, or 6, or 9;

    (c)  2B - 3C >= 12.


From (c), if C= 0, then admittable values for B are 6 or 9;

          if C= 2, then admittable value for B is 9;

             C can not be 4, or 6, or 8  ( from inequalities (c) )


So, the only possible blocks of two digits BC can be 60, 90, 92.


If BC is 60, then from (3)  12*A - 2*6 + 3*0 = 0,  hence,  A = 1.  So, AC = 10, ABC = 160.

If BC is 90, then from (3)  12*A - 2*9 + 3*0 = 0,  hence, A = 18/12 = 1.5, which is not an integer number, so this case does not work.

If BC is 92, then from (3)  12*A - 2*9 + 3*2 = 0,  hence, A = 12/12 = 1.  So, AC = 12, ABC = 192.


Thus, the maximum possible number ABC is 192.    


ANSWER.  The only maximum possible number ABC under given condition is 192.