Let two digit number be $ab$, it's actual value is $10a + b$
$a + b = 12$ (sum of digits is $12$)
Number formed by reversing the digits is greater than the original number by $54$
$\Rightarrow 10b +a - (10a + b) = 54$
$\Rightarrow b -a = 6$
Solving $a+b=12$ and $b - a = 6$, We get $a = 3,b = 9$
Thus $39$ is the required number.
