We will have the following:
We recall that the range of a projectile is given by:
[tex]R=\frac{v_i^2sin(2\theta)}{g}[/tex]Then, we will have that the range of the flare will be:
[tex]\begin{gathered} R=\frac{(190m/s)sin(2\ast45)}{(9.8m/s)}\Rightarrow R=\frac{180500}{49}m \\ \\ \Rightarrow r\approx3683.67m \end{gathered}[/tex]So, the flare will land approximately 3 683.67 meters away.