Given differential equation

$\frac{dp}{dt}=0.5p-450$

with initial condition

$p(0)=850.$

Rewrite the equation

$\frac{dp}{dt}-0.5p=-450.$

This is a linear differential equation.

Find the general solution

For $\frac{dp}{dt}-0.5p=-450,$ the steady-state value is obtained by setting $\frac{dp}{dt}=0$:

$0.5p-450=0 \implies p=900.$

So write the solution as

$p(t)=900+Ce^{0.5t}.$

Check: $\frac{dp}{dt}=0.5Ce^{0.5t},$ and $0.5p-450=0.5(900+Ce^{0.5t})-450=450+0.5Ce^{0.5t}-450=0.5Ce^{0.5t},$ which matches.

Use the initial condition

$p(0)=900+C=850$ $C=-50.$

Hence,

$p(t)=900-50e^{0.5t}.$

Find when population becomes zero

Set $p(t)=0$:

$900-50e^{0.5t}=0$ $50e^{0.5t}=900$ $e^{0.5t}=18.$

Taking natural logarithm,

$0.5t=\ln 18$ $t=2\ln 18.$

Compare with options

The correct option is:

$\boxed{A:\;2\ln 18}$