The bullet's position vector has components
[tex]x=v_0t[/tex]
[tex]y=-\dfrac g2t^2[/tex]
where we're given the initial velocity [tex]v_0[/tex] and the initial height of the bullet [tex]y_0[/tex], and [tex]g[/tex] is the acceleration due to gravity. Note that we take the bullet's starting position to be the origin.
We're also told that after some time [tex]t[/tex], the time it takes for the bullet to hit the target, the bullet's vertical position is 0.64 m below where it was aimed, which translates to [tex]y=-0.64\,\mathrm m[/tex] at this time [tex]t[/tex]. Then we can solve for this time:
[tex]-0.64\,\mathrm m=-\dfrac{9.80\,\frac{\mathrm m}{\mathrm s^2}}2t^2\implies t=0.36\,\mathrm s[/tex]
Finally, the target's distance from the origin is [tex]x[/tex] at this time [tex]t[/tex]:
[tex]x=\left(154.6\,\dfrac{\mathrm m}{\mathrm s}\right)(0.36\,\mathrm s)=55.87\,\mathrm m[/tex]
