Answer
$m'= 0.13 \space mg$
Work Step by Step
The original amount of $^{238}U$ is
$m_o = m e^{\lambda t}$,
$m_o = 3.7 \times 10^{-3}g.e^{ln2 \times (260 \times 10^6 y/ 4.47 \times 10^9 y)}$
$m_o = 3.85 \times 10^{-3} g$
$m' = (m_o - m) [\frac{m_{Pb}}{m_{U}}]$
$m' = (3.85 \times 10^{-3} g - 3.70 \times 10^{-3} g ) [\frac{206}{238}]$
$m'=0.132 \times 10^{-3} g$
$m'= 1.3 \times 10^{-4}g$
$m'= 0.13 \space mg$
