The position vector of the bullet has components
[tex]x=v_0t[/tex]
[tex]y=1.4\,\mathrm m-\dfrac g2t^2[/tex]
The bullet hits the ground when [tex]y=0[/tex], which corresponds to time [tex]t[/tex]:
[tex]1.4\,\mathrm m-\dfrac g2t^2=0\implies t=0.53\,\mathrm s[/tex]
The bullet travels 168 m horizontally, which would require a muzzle velocity [tex]v_0[/tex] such that
[tex]168\,\mathrm m=v_0(0.53\,\mathrm s)[/tex]
[tex]\implies v_0\approx320\,\dfrac{\mathrm m}{\mathrm s}[/tex]
