Let V be volume of body using law of floatation.
In the first case :
The upthrust =weight of body
$W= \large\frac{1}{3} $$V \rho _{Hg}g$-------(1)
In the second case :
Let x fraction of the volume be immersed in mercury.
Then $W=x V \rho _{Hg}+(1-x) V \rho _{liquid}g$--------(2)
equating (1) and (2)
$\large\frac{\rho _{Hg}}{3}$$=x \rho _{Hg}+(1-x) \rho _{liquid}$
$\large\frac{\rho _{Hg}}{3}$$=x \rho _{Hg}+(1-x) \large\frac{\rho _{Hg}}{10}$
$\large\frac{1}{3}$$=x +\bigg(\large\frac{1-x}{10}\bigg)$
$\large\frac{1}{3}$$=x -\large\frac{x}{10}+\frac{1}{10}$
$\large\frac{1}{3}-\frac{1}{10}=\frac{9}{10}$$x$
$\large\frac{10-3}{30}=\frac{9}{10}$$x$
$\large\frac{7}{3}$$=9x$
$x= \large\frac{7}{27}$$=0.25$
Hence b is the correct answer.