The average force exerted on the bullet can be calculated by using Second Newton Law:
[tex]F_{\text{avg}}=ma[/tex]where m is the mass of the bullet and a its acceleration. Consider that acceleration is determined by the following expression (from kinematics):
[tex]a=\frac{v-v_o}{t}[/tex]where v is the final speed of the bullet, vo the initial speed and t the time of the change of the velocity.
Then, by replacing the previous expression for a into the formula for Favg, you have:
[tex]F_{\text{avg}}=m(\frac{v-v_o}{t})[/tex]In this case,
m = 0.0100 kg
v = 450m/s
vo = 0m/s (the bullet starts from rest)
t = 1.80ms = 1.80*10^-3 s
Replace the previous values of the parameters into the formula for Favg and simplify:
[tex]\begin{gathered} F_{\text{avg}}=(0.0100kg)(\frac{450\frac{m}{s}-0\frac{m}{s}}{1.80\cdot10^{-3}s}) \\ F_{\text{avg}}=2500N \end{gathered}[/tex]Hence, the average force exerted on the given bullet is 2500N