Answer
$M_{Rn} = 6.425 \times 10^{-9} g$
Work Step by Step
$R_{Ra} = R_{Rn}$ and $R = \lambda N = (ln 2/T_{1/2})(M/m)$
$\frac{ln 2}{T_{1/2Rn}} . \frac{M_{Rn}}{m_{Rn}} = \frac{ln 2}{T_{1/2Ra}} . \frac{M_{Ra}}{m_{Ra}} $
$M_{Rn} = [\frac{T_{1/2Rn} ln2 }{T_{1/2Ra} ln 2}][\frac{m_{Rn}}{m_{Ra}}] M_{Ra} $
Cancel out ln 2, we get
$M_{Rn} = [\frac{T_{1/2Rn} }{T_{1/2Ra}}][\frac{m_{Rn}}{m_{Ra}}] M_{Ra} $
Plug in all the values
$M_{Rn} = [\frac{3.82 d}{1600y \times 365 d/y}][\frac{1.0 \times 10^{-3}g \times 222u}{226u}]$
$M_{Rn} = 6.425 \times 10^{-9} g$
