Exponentil bacteria growth 

a) Let the exponential model for the number of bacterias in a culture after $t$ years will be of the form
$$N=ae^{bt}$$
Using the initial condition at time zero, number of bacterias are 300, so
$$300=ae^0$$
$$\implies a=300$$

and after 9 years number of bacterias are doubled, so
$$2\times 300=300e^{b\times 9}$$
$$2=e^{9b}$$
Taking antilogarithm we get
$$\ln 2=9b$$
or $$b=\frac19 \ln2$$
So, the model is
$$N=300e^{\frac{\ln2 }{9}t}$$

b) The time it will require to grow bacteria to 1000 will be given by
$$1000=300e^{\frac{\ln2 }{9}t}$$
$$\frac{10}{3}=e^{\frac{\ln2 }{9}t}$$
Taking antilogarithm we get
$$\ln(\frac{10}3)=\frac{\ln2}{9}t$$
So, the time required is 
$$t=\frac{9\ln(\frac{10}3)}{\ln2}\approx 15.6327\;\;\text{ years}$$