Mass of the planet is M. If the mass of the planet reduced to ${\text{1/8}}^{\text{th}}$  of the original without change in density. Then what is the new value of acceleration due to gravity of the planet

$\text{g}=\frac{\text{GM}}{{\text{R}}^2}$ $=\frac{\text{G}\frac{4}{3}\pi {\text{R}}^3\rho }{{\text{R}}^2}$ $\Rightarrow \text{g}=\text{G}\frac{4}{3}\pi \text{R}\rho$ 
 

Since the density is constant, we can say:
$\text{g}\propto \text{R}$                            

$\frac{{\text{g}}_{\text{1}}}{{\text{g}}_{\text{2}}}=\frac{{\text{R}}_{\text{1}}}{{\text{R}}_{\text{2}}}$            

Mass is changed without changing the density. We don't know how the radius of the planet is changed.That can be found from the following:
$\text{M}=\frac{4}{3}\pi {\text{R}}_{\text{1}}^{\text{3}}\rho$$,\frac{\text{M}}{8}=\frac{4}{3}\pi {\text{R}}_2^{\text{3}}\rho$

$\frac{\text{M}}{\frac{\text{M}}{8}}=\frac{\frac{4}{3}\pi {\text{R}}_{\text{1}}^{\text{3}}\rho }{\frac{4}{3}\pi {\text{R}}_2^{\text{3}}\rho }$

$8=\frac{{\text{R}}_{\text{1}}^{\text{3}}}{{\text{R}}_2^{\text{3}}}$ $\Rightarrow 2=\frac{{\text{R}}_{\text{1}}}{{\text{R}}_{\text{2}}}$

$\frac{{\text{g}}_{\text{1}}}{{\text{g}}_{\text{2}}}=2\Rightarrow {\text{g}}_{\text{2}}=\frac{{\text{g}}_{\text{1}}}{2}$