It could be 0, because the sum of its digit is 0 and 0 is 13x0.
But that's stretching it a little. 0 is the only 1-digit number
that could be a solution because no other digit is a multiple of 13.

It couldn't be a 2-digit number because the only 2-digit 
multiples of 13 are 13,26,39,52,65,78,91 and none of those 
are 13 times their sum of digits.

It can't be a 4 digit number or larger because the largest
4-digit number 9999 has sum of digits of only 36, and 36x13
is only 468, not a 4-digit number.  

Therefore it has to be a three digit number.

Let h = its hundreds digit
Let t = its tens digit
Let u = its units digit

100h+10t+u = 13(h+t+u)
100h+10t+u = 13h + 13t + 13u
       87h = 3t + 12u
       29h = t + 4u

The largest t + 4u could be would be when t and u are both 9.
making t + 4u = 9 + 4(9) = 45. thus

       29h <= 45
         h <= 

h can only be 1, and 29h = t + 4u becomes

        29 = t + 4u

Solving for t

     29 - 4u  = t

      -1 < t < 10

   -1 < 29 - 4u < 10
  -30 <   -4u   < -19
   30 >    4u   >  19

Divide through by 4

   7.5 > u > 4.75

Therefore u is 5, 6 or 7, the only digits in that interval.

If u = 5, substitute in

29 - 4u  = t
29 - 4(5) = t
29 - 20 = t
      9 = t

So one solution is h=1, t=9, u=5, or the number 195

If u = 6, substitute in

29 - 4u  = t
29 - 4(6) = t
29 - 24 = t
      5 = t

Another solution is h=1, t=5, u=6, or the number 156 

If u = 7, substitute in

29 - 4u  = t
29 - 4(7) = t
29 - 28 = t
      1 = t

Another solution is h=1, t=1, u=7, or the number 117
 
There are three solutions 117, 156, and 195.  And you might
mention that 0 could also be considered as a solution, but it's 
a little far-fetched.

Edwin