If the Earth shrinks such that its density becomes 8 times to the present value, then the new during of the day in hours will be

${\mathrm{d}}_2=8\mathrm{d}, \Rightarrow \frac{{\mathrm{d}}_2}{{\mathrm{d}}_1}=8$

We know, mass$\mathrm{M}=\frac{4}{3}{\mathrm{\pi R}}^3\mathrm{d}\Rightarrow \mathrm{d}\propto \frac{1}{{\mathrm{R}}^3}$

${\left(\frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}\right)}^3=\frac{{\mathrm{d}}_2}{{\mathrm{d}}_1}=8\Rightarrow \frac{{\mathrm{R}}_1}{{\mathrm{R}}_2}=2$

From conservation of angular momentum :

${\mathrm{I}}_1{\mathrm{\omega }}_1={\mathrm{I}}_2{\mathrm{\omega }}_2$

$\Rightarrow \frac{2}{3}{{\mathrm{MR}}_1}^2\times \frac{2\mathrm{\pi }}{{\mathrm{T}}_1}=\frac{2}{5}{{\mathrm{MR}}_2}^2\times \frac{2\mathrm{\pi }}{{\mathrm{T}}_2}$

$\frac{{\mathrm{T}}_2}{{\mathrm{T}}_1}={\left(\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}\right)}^2\Rightarrow \frac{{\mathrm{T}}_2}{24}={\left(\frac{1}{2}\right)}^2\Rightarrow {\mathrm{T}}_2\frac{24}{4}=6\mathrm{hr}$