Answer
The correct drawing is (d).
Work Step by Step
The gravitational force exerted on object 3 by object 1 is $$F_{13}=G\frac{m_1m_3}{r_{13}^2}$$
The gravitational force exerted on object 3 by object 2 is $$F_{23}=G\frac{m_2m_3}{r_{23}^2}$$
$F_{13}$ and $F_{23}$ are in opposite direction. For the net gravitational force on object 3 to be zero, $$\sum F_3=F_{13}-F_{23}=G\frac{m_1m_3}{r_{13}^2}-G\frac{m_2m_3}{r_{23}^2}=0$$ $$\frac{m_1}{r_{13}^2}-\frac{m_2}{r_{23}^2}=0$$ $$m_1r_{23}^2-m_2r_{13}^2=0$$ $$m_1r_{23}^2=m_2r_{13}^2$$
We know that $m_2\gt m_1$. For the equation to be equal, it is necessary that $r_{13}^2\lt r_{23}^2$ and thus, $r_{13}\lt r_{23}$
Therefore, object 3 needs to be closer to object 1 than object 2. (d) is the correct drawing.
