Answer
$0.0835\,\Omega$
Work Step by Step
Recall that $(I-I_{g})R=I_{g}r$
where $I_{g}$ is the full-scale galvanometer current, $I$ is the full-scale ammeter current, $r$ is the internal resistance of galvanometer and $R$ is the shunt resistance.
Then $(60.0\times10^{-3}\,A-0.100\times10^{-3}\,A)R=(0.100\times10^{-3}\,A)(50.0\,\Omega)$
$R=\frac{(0.100\times10^{-3}\,A)(50.0\,\Omega)}{(60.0\times10^{-3}\,A-0.100\times10^{-3}\,A)}=0.0835\,\Omega$
