In the reported figure of earth, the value of acceleration due to...

${g}_{A} = \dfrac{GM\left(r\right)}{{R}^{3}}$

${g}_{C} = \dfrac{GM}{{\left(R + \dfrac{R}{2}\right)}^{2}}$
${g}_{A} = {g}_{C}$

$\dfrac{r}{{R}^{3}} = \dfrac{1}{\dfrac{9}{4}{R}^{2}} ⇒ r = \dfrac{4R}{9}$

so $OA = \dfrac{4R}{9}; AB = R - r = \dfrac{5R}{9}$

$OA: AB = 4: 5$