The digit at the tens place …

Let the digits at unit place be y. Let the digits at ten's place be x.
Number = 10x + y
Sum of digits of the two digits number is four times that in the unit's place. => x + y = 4y => x = 4y - y => x = 3y ......(i).
If the digits are reversed the new number will be 54 less that the original number.
Number obtained by reversing the digits = 10y + x
Number obtained by reversing the digits = 10x + y - 54 => 10y + x = 10x + y - 54 => 54 = 10x + y - 10y - x => 54 = 9x - 9y => 54 = 9(x - y) => 54/9 = x - y => 6 = x - y .........(ii).
Putting the value of x from Eq (i). in Eq (ii). => 6 = x - y => 6 = 3y - y => 6 = 2y => y = 6/2 => y = 3
Putting the value of y in Eq (ii). => 6 = x - y => 6 = x - 3 => -x = -6 - 3 => -x = -9 => x = 9
So, Number = 10x + y => 10(9) + 3 => 90 + 3 => 93
Hence, the original number is 93.