Answer
$a=0 \ T +|a| \ N$
Work Step by Step
$v(t)=\dfrac{dr}{dt}=(-a \sin t) i +(a \cos t) j +b \ k \implies |v(t)|=\sqrt {a^2+b^2}$
and $a(t)=\dfrac{d \ v(t)}{dt}=(-a \cos t) i +(-a \sin t) j + 0 \ k$
$|a(t)|=\sqrt {a^2(\cos^2 t+\sin ^2 t)} \implies |a(t)|=a$
Now, $a_{N}=\sqrt {|a|^2 -a^2_{T}}=\sqrt {a^2-0}=|a|$
So, $a=a_T T+a_{N}=0 \ T +|a| \ N$