How to apply the Heisenberg Uncertainty Principle in calculations on the movement of bacteria?
$\newcommand{\upmu}{{\large\unicode[Times]{x3BC}}}%Remove this line when upright greek characters are implemented by mathjax$
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The velocity of the  bacteria is $\mathbf{v} = 9~\mathrm{\upmu m/s} = 9\times10^{-6}~\mathrm{m/s}$, with an uncertainty of $9.00\%$ you can calculate the absolute uncertainty in the velocity $\Delta \mathbf{v}$ as the difference of the upper bound $(\mathbf{v} + 0.09\mathbf{v})$ and the lower bound $(\mathbf{v} - 0.09\mathbf{v})$.


   $$\Delta \mathbf{v} = (\mathbf{v} + 0.09\mathbf{v}) - (\mathbf{v} - 0.09\mathbf{v}) = 0.18\mathbf{v} = 1.62\times10^{-6}~\mathrm{m\,s^{-1}}$$


The momentum is defined as 
$$\mathbf{p} = m\mathbf{v},$$ 
hence the absolute uncertainty in the momentum is 
$$\Delta\mathbf{p} = m\Delta\mathbf{v}.$$
The mass is $m = 0.100~\mathrm{fg} = 1.00\times10^{-19}~\mathrm{kg}$.


   $$\Delta\mathbf{p} = m\Delta\mathbf{v} = 1.62\times10^{-25}~\mathrm{kg\,m\,s^{-1}}$$


Now you can use this to calculate the uncertainty in position $\Delta\mathbf{x}$.
$$\Delta\mathbf{x}\Delta\mathbf{p} \geq \frac{h}{2}$$
Planck's constant is $h = 6.62607004\times 10^{-34}~\mathrm{m^2\,kg\,s^{-1}}$.


   $$\Delta\mathbf{x} \geq \frac{h}{2\Delta\mathbf{p}}\\\Delta\mathbf{x} \geq 2.07\times10^{-9}~\mathrm{m}$$


Since you have not included any of the steps how you arrived at your result, I cannot help you finding your mistake.