The weight of lump is given as,

$W _1= mg$

The volume of a lump is given as,

$V =\frac{ m }{\rho_1}$

The weight of lump in brine solution is given as,

$W _2= mg -\frac{\rho g V }{2}$

$= mg \left(1-\frac{\rho}{2 \rho_1}\right)$

The new readings of the spring balance is given as,

$W^{\prime}=m g\left(1-\frac{\rho}{2 p_1}\right)$

$=200 \times\left(1-\frac{1.1}{2 \times 11.4}\right)$

$=190.35\,gF$

Thus, the new reading of the spring balance is $190.35\,gF$.