Answer
$1.5\times 10^9 \;\rm m$
Work Step by Step
First, we need to sketch the force diagram exerted on our satellite by the Earth and the Sun, as shown below.
According to the author, we need to find $r$ in the figure below.
The net force exerted on the satellite is given by Newton's second law;
$$\sum F_r=F_{\text{Sun on Satellite}}-F_{\text{Earth on Satellite}}=m_{\rm Satellite}a_r$$
Applying Newton's gravitational law;
$$ \dfrac{Gm_{\rm Sun}\color{red}{\bf\not}m_{\rm Satellite}}{R_s^2}-\dfrac{Gm_{\rm Earth}\color{red}{\bf\not}m_{\rm Satellite}}{r^2}=\color{red}{\bf\not}m_{\rm Satellite}\dfrac{v^2}{R_s}$$
$$ \dfrac{Gm_{\rm Sun}} {R_s^2}-\dfrac{Gm_{\rm Earth} }{r^2} =\dfrac{v^2}{R_s}$$
where $v=2\pi R_s/T$ and the period of our satellite is the same as the Earth, one year.
$$ \dfrac{Gm_{\rm Sun}} {R_s^2}-\dfrac{Gm_{\rm Earth} }{r^2} =\dfrac{4\pi^2 R_s^{\color{red}{\bf\not}2}}{T^2\color{red}{\bf\not}R_s}$$
where $R_s=R_E-r$
$$ \dfrac{Gm_{\rm Sun}} {(R_E-r)^2}-\dfrac{Gm_{\rm Earth} }{r^2} =\dfrac{4\pi^2 (R_E-r)}{T^2 }\tag 1$$
where $T=T_e$ and is given by
$$\dfrac{GM_{\rm Sun}\color{red}{\bf\not}m_{\rm Earth}}{R_E^{\color{red}{\bf\not}2}}=\color{red}{\bf\not}m_{\rm Earth}\dfrac{v^2}{\color{red}{\bf\not}R_E}$$
$$\dfrac{GM_{\rm Sun} }{R_E }= \dfrac{4\pi^2 R_E^2}{T^2 }$$
Thus, $$T^2=\dfrac{4\pi^2 R_E^3}{GM_{\rm Sun} }$$
Plugging into (1);
$$ \dfrac{\color{red}{\bf\not}Gm_{\rm Sun}} {(R_E-r)^2}-\dfrac{\color{red}{\bf\not}Gm_{\rm Earth} }{r^2} =\dfrac{\color{red}{\bf\not}4\color{red}{\bf\not}\pi^2 (R_E-r)\color{red}{\bf\not}Gm_{\rm Sun} }{ \color{red}{\bf\not}4\color{red}{\bf\not}\pi^2 R_E^3 } $$
Now we need to solve for $r$ and the author told us to use the binomial approximation.
$$ \dfrac{ m_{\rm Sun}} {(1-\frac{r}{R_E})^2R_E^2}-\dfrac{ m_{\rm Earth} }{r^2} =\dfrac{ (1-\frac{r}{R_E})\color{red}{\bf\not}R_E\;m_{\rm Sun}}{R_E^{\color{red}{\bf\not}3} }$$
$$ \dfrac{ m_{\rm Sun}(1-\frac{r}{R_E})^{-2}} {R_E^2}-\dfrac{ m_{\rm Earth} }{r^2} =\dfrac{ (1-\frac{r}{R_E}) m_{\rm Sun}}{R_E^2 }$$
where $[r/R_E\lt \lt 1]$, so $(1-\frac{r}{R_E})^{-2
}\approx 1-(-2
)\frac{r}{R_E}$;
$$ \dfrac{ m_{\rm Sun}(1+2\frac{r}{R_E}) } {R_E^2}-\dfrac{ m_{\rm Earth} }{r^2} =\dfrac{ (1-\frac{r}{R_E}) m_{\rm Sun}}{R_E^2 }$$
$$ \dfrac{ m_{\rm Sun}(1+2\frac{r}{R_E}) } {R_E^2}- \dfrac{ (1-\frac{r}{R_E}) m_{\rm Sun}}{R_E^2 }=\dfrac{ m_{\rm Earth} }{r^2} $$
$$ \dfrac{ m_{\rm Sun}\left[(1+2\frac{r}{R_E})- (1-\frac{r}{R_E}) \right]}{R_E^2 }=\dfrac{ m_{\rm Earth} }{r^2} $$
$$ \dfrac{ m_{\rm Sun}\left[ 1+2\frac{r}{R_E} - 1+\frac{r}{R_E} \right]}{R_E^2 }=\dfrac{ m_{\rm Earth} }{r^2} $$
$$ \dfrac{ 3rm_{\rm Sun} }{R_E^3}=\dfrac{ m_{\rm Earth} }{r^2} $$
Thus,
$$r^3=\dfrac{ m_{\rm Earth} R_E^3}{3 m_{\rm Sun} } $$
$$r =\sqrt[3]{\dfrac{ m_{\rm Earth} R_E^3}{3 m_{\rm Sun} } }$$
$$r =R_E\cdot \sqrt[3]{\dfrac{ m_{\rm Earth} }{3 m_{\rm Sun} } }$$
Plugging from table 13.2;
$$r =(1.5\times 10^{11})\cdot \sqrt[3]{\dfrac{ (5.98\times 10^{24}) }{3 (1.99\times 10^{30}) } }=\color{red}{\bf1.5\times 10^9}\;\rm m$$
