The charge of an electron is $1.6\times {10}^{-19}$ C. How many electrons pass through the cross section of a copper wire carrying ${10}^{-16}$ amp  in 1 second ?

Here we have,

${\text{e}}^{-}=1.6\times {10}^{-19}C$, $\mathrm{current} \mathrm{i} \mathrm{is}={10}^{-16}\mathrm{A} \mathrm{for} \mathrm{t}=1\sec$

We know that charge $\mathrm{q}=\mathrm{i}\times \mathrm{t}={10}^{-16}\times 1={10}^{-16}\mathrm{ampsec}\left(\mathrm{C}\right)$

 $\mathrm{Number} \mathrm{of} {\mathrm{e}}^{-} \mathrm{passing} \mathrm{per} \sec =\frac{{10}^{-16}}{1.6\times {10}^{-19}}=\frac{{10}^3}{1.6}=625$

Therefore option D is correct.