The gravitational attraction between two bodies is calculated as follows:

#F=(G*m_1*m_2)/d^2#

with #G# the universal gravity constant, #m_1# and #m_2# the masses of the two bodies and #d# the distance between the two bodies.

If we place a body on a straight line between the earth and the sun, the resulting gravitational attractions will be:

#F_(sb)=(G*m_s*m_b)/((d_(sb))^2)#

#F_(eb)=(G*m_e*m_b)/((d_(eb))^2)#

We are considering the situation when #F_(sb)=F_(eb)#:

#(cancel(G)*m_s*cancel(m_b))/((d_(sb))^2)=(cancel(G)*m_e*cancel(m_b))/((d_(eb))^2)#

#rarr (m_s)/((d_(sb))^2)=(m_e)/((d_(eb))^2)#

#rarr (m_s)/(m_e)=((d_(sb))^2)/((d_(eb))^2)=((d_(sb))/(d_(eb)))^2#

#rarr sqrt((m_s)/(m_e))=(d_(sb))/(d_(eb))#

Knowing that #m_e=6*10^24#kg and #m_s=2*10^30#kg, and that #d_(sb)+d_(eb)=1.5*10^11#m we have to solve the following equation:

#sqrt((2*10^30)/(6*10^24))=(1.5*10^11-d_(eb))/(d_(eb))#

#rarr d_(eb)=(1.5*10^11-d_(eb))/(sqrt((2*10^30)/(6*10^24)))=(1.5*10^11-d_(eb))/(sqrt(1/3*10^6))=((1.5*10^11-d_(eb))*sqrt3)/(10^3)#

#rarr d_(eb)*10^3+sqrt(3)*d_(eb)=1.5sqrt3*10^11#

#rarr d_(eb)(10^3+sqrt3)=sqrt6.75*10^11#

#rarr d_(eb)=(sqrt6.75*10^11)/(10^3+sqrt3)~~259358400#m