Answer
$N_U = 1.06 \times 10^{19}$
Work Step by Step
The mass of a single $^{238}U$ atom is,
$m_U = 238 u \times 1.661 \times 10^{-24} g/u $
$m_U = 3.95 \times 10^{-22} g$
so the number of uranium atom in $4.20 \times 10^{–3} g$ is
$N_U = \frac{4.20 \times 10^{–3} g}{3.95 \times 10^{-22} g}$
$N_U = 1.06 \times 10^{19}$
