Answer:
Here, \[g=\frac{GM}{{{R}^{2}}}=\frac{G}{{{R}^{2}}}.\frac{4}{3}\pi {{R}^{3}}\rho =\frac{4}{3}\pi GR\rho \] or \[g\propto R\rho \] \[\therefore \] \[\frac{{{g}_{1}}}{{{g}_{2}}}=\frac{R\rho }{2R.\frac{\rho }{2}}=\mathbf{1:1}\mathbf{.}\]
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