Denote the number by 10a + b, where both a and b are integers picked from {1, 2, 3, …, 9}.
"the difference of whose digits is 3" ==> |a - b| = 3
"4 times the number is equal to 7 times the number obtained by reversing the digits" ==> 4 (10a + b) = 7 (10b + a)
Simplifying the second equation, you get
40a + 4b = 70b + 7a
33a - 66b = 0
a - 2b = 0
Assume a > b, in which case we would have |a - b| = a - b, so
a - b = 3
Eliminate a to solve for b, then for a :
(a - b) - (a - 2b) = 3 - 0
b = 3 ==> a = 6
Then the original number is 63.
If instead we had assumed a < b, we would have had |a - b| = b - a = 3. Then
2 (b - a) + (a - 2b) = 2×3 + 0
-a = 6
But neither a nor b can be negative, so this case is moot.