If k is the current sensitivity of the galvanometer the current in the galvanometer is
$l_g=50$ K
when a shunt S is connected across it the current through the galavnometer becomes
$l_g^{\prime }=\frac{l_{g}S}{G+S}=10$ K
where G is the resistance of the galvanometer dividing the equations we get
$\frac{G+S}{S}=5$
which gives $G=4\times S=4\times 12=48$ ohm