Answer
$v_S=4.61\frac{m}{s}$
Work Step by Step
According to the Doppler effect
$f^-=f\frac{v+_-v_D}{v+_-v_S}$
In the current scenario, the source (ambulance) is approaching the detector (cyclist) so we will use $+$ sign.
$f^-=f\frac{v+v_D}{v+v_S}$
Given that
$v_D=2.44\frac{m}{s}$
$v=343\frac{m}{s}$
$f^-=1590Hz$ , $f=1600Hz$
$f^-=f\frac{v+v_D}{v+v_S}$
$v+v_S=\frac{f}{f^-}(v+v_D)$
$v_S=\frac{f}{f^-}(v+v_D)-v$
putting the values, we get
$v_S=\frac{1600}{1590}(343+2.44)-343=4.61\frac{m}{s}$
Thus, the speed of the ambulance is
$v_S=4.61\frac{m}{s}$
