time $t$ when quantity is double in differential equation

The evolution of certain quantity Represented by differential equation $$\displaystyle \frac{dz}{dt}=1.5z.$$ How long it take for $$z$$ to be double. If time $$t$$ measure in hours.

What i Try: $$\frac{dz}{z}=1.5dt\Longrightarrow \int^{2z_{0}}_{z_{0}}\frac{dz}{z}=\int^{t}_{0}1.5dt$$

$$\ln(2)=1.5t\Longrightarrow t=3\ln(2)$$

I have a doubt. When time $$t=0.$$ Can we take population $$z_{0}$$.

Please tell is is my thinking is right. Also please tell me is my solution is correct. Thanks

I think your confusion results from a combination of imprecise/inconsistent notation (e.g. wanting to choose a general $$z_0$$ while fixing $$t_0 = 0$$) and attempting to use a definite integral to solve the equation (when a general solution is clearer and more precise). I think a more instructive way to write the same work is as follows:
We first deduce a general solution to the differential equation, \begin{align*} \frac{\mathrm{dz}}{\mathrm{dt}} = 1.5z &\Leftrightarrow \frac{\mathrm{dz}}{z} = 1.5 \mathrm{dt} \\ &\Leftrightarrow \int \frac{\mathrm{dz}}{z} = \int 1.5 \mathrm{dt} \\ &\Leftrightarrow \mathrm{ln}(z) = 1.5t + C \\ &\Leftrightarrow z = D e^{1.5t}, \end{align*} so that if we have $$z_0 = De^{1.5t_0}$$ and $$2z_0 = De^{1.5t_1}$$ are quantities occurring at distinct times $$t_0$$ and $$t_1$$, then \begin{align*} \frac{2z_0}{z_0} &= \frac{De^{1.5t_1}}{De^{1.5t_0}} \\ 2 &= e^{1.5(t_1-t_0)}. \end{align*} That is, the doubling time is $$t_1 - t_0 = \frac{\mathrm{ln}(2)}{1.5}$$ (note, as SarGe mentions, that your answer has a small error here).
Writing things this way emphasizes the independence of the doubling time from the "free variable" $$D$$ (as $$D$$ cancels in the calculations), which is an important conceptual idea. In addition, that our final answer is a difference $$t_1 - t_0$$ (not a single value $$t$$) reinforces that for this differential equation the doubling time has nothing to do with the exact starting/stopping values, just their difference.
Yes, you can take initial population to be $$z_0$$ or anything else provided the final population is double of that. Also the general relation between population and time is given as $$\ln \left(\frac{z}{z_0} \right) =1.5t$$ Hence, there is correction in your answer, it should be $$\displaystyle\frac{2}{3}\ln 2$$.