Explanation
Each dipole has equal and opposite charge. So net charge of dipole is zero.
Thus the total charge inside the sphere is zero. According to Gauss's law, the flux, $$\phi=\dfrac{q_{in}}{\epsilon_0}=0 $$ as charge inside sphere is zero.
$$\int { dV } =-\int { E.dr } \Rightarrow \quad \int _{ { V }_{ 0 } }^{ V }{ dV } =-\int _{ 0 }^{ Z }{ \dfrac { \sigma }{ 2{ \in }_{ 0 } } dZ } $$
$$\Rightarrow V-{ V }_{ 0 }=-\dfrac { \sigma Z }{ 2{ \in }_{ 0 } } \Rightarrow V={ V }_{ 0 }-\dfrac { \sigma Z }{ 2{ \in }_{ 0 } } $$
Here, $$F_{air}=\dfrac{q_1q_2}{4\pi\epsilon_0 d^2_{air}}$$ and
$$F_{oil}=\dfrac{q_1q_2}{4\pi\epsilon_r\epsilon_0 d^2_{oil}}$$
According to question, $$F_{air}=F_{oil} $$
or $$\dfrac{q_1q_2}{4\pi\epsilon_0 d^2_{air}}=\dfrac{q_1q_2}{4\pi\epsilon_r\epsilon_0 d^2_{oil}}$$
or $$ \epsilon_r d^2_{oil}=d^2_{air}$$
or $$d^2_{oil}=\dfrac{(50)^2}{5}=500$$
or $$d_{oil}=22.3 m$$
Here, $$r=\dfrac{3}{2} =1.5m$$ and charges are $$q_1=20$$ and $$q_2=6$$
Coulomb's law,
$$F=k\dfrac{q_1q_2}{r^2}$$
Therefore, (for simplicity assume k=1)
$$F=\dfrac{20 \times 6}{(1.5)^2}=53.3N$$
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