#### Answer

The Rutheford's foil is $1.7 \times 10^4$ atoms thick.

#### Work Step by Step

$1$ $mm$ = $10^{-3}$ $m$
$0.005$ $mm$ = $0.005$ $\times 10^{-3}$ $m$ = $5 \times 10^{-6}$ $m$ = Foil thickness.
Thus, it follows:
$\frac{5 \times 10^{-6}m}{Foil} \times \frac{1-atom}{2.9 \times 10^{-10}m} = 1.7 \times 10^4 \frac{Atoms}{Foil}$