Isn't there multiple answers for this?

$(1000a+100b+10c+d) + (a+b+c+d) = 2010$ and $a+b+c+d>11$ So $a=1$ and $101b+11c+2d=1009 \to 101b \geq 1009-9*11-2*9 \to b =9$ $11c+2d=100 \to 11c \geq 100-2*9 \to c=8$ and $d=6$ and so he born in $1986$ and is $24$ years old Case 2: If he didn`t have birthday in this year, then $(1000a+100b+10c+d)+(a+b+c+d)=2009$ Same way $a=1,b=9$ $11c+2d=99 \to c=9,d=0$ And so he born in $1990$ and is $19$ years old