Answer
$40\;cm$
Work Step by Step
For first object, the equilibrium occurs 10 cm below $y_i$
Therefore, at equilibrium
$100g=k\times10$
or, $k=10g\;dyn/cm$
Let, the new equilibrium (rest) position with both objects attached to the spring is $x\;cm$ below $y_i$
Thus
$kx=(300+100)g$
or, $10g\times x=400g$
or, $x=40\;cm$
