We are not given tables, so will just use the amortization formula.
[tex]P=\frac{A*((1+i)^n-1)}{i*(1+i)^n}[/tex]
where
P=amount to be deposited today, to be found
A=amount withdrawn each year=18000
i=Annual interest=9%
n=number of years = 20
Substituting values,
[tex]P=\frac{A*((1+i)^n-1)}{i*(1+i)^n}[/tex]
[tex]=\frac{18000*((1+0.09)^{20}-1)}{0.09*(1+0.09)^{20}}[/tex]
=164313.82 to the nearest cent
