Force of gravitation on \(m_0\) due to \(m=\frac{G m m_0}{r^2}=F_1\)
Force of gravitation on \(m_0\) due to \(4 m=\frac{G 4 m m_0}{(d-r)^2}=F_2\)
Net force \(=0\)
\(\Rightarrow F_1=F_2\)
\(\frac{G m m_0}{r^2}=\frac{4 G m m_0}{(d-r)^2}\)
\(\Rightarrow (d-r)^2=(2 r)^2\)
\(\Rightarrow d-r=2 r\)
\(\Rightarrow d=3 r\)
Thus, \(r=\frac{d}{3}\)