$\xrightarrow[\bar{I}=500\mu A=50\times {{10}^{-6}}A]{\Delta t=400h}\Delta q=(500\times {{10}^{-6}})\times 400$   $\to \Delta q=2\times {{10}^{-1}}Ah\times \frac{{{10}^{3}}mA}{1A}=2\times {{10}^{2}}mAh$