Two thin uniform rings of mass ${\mathrm{m}}_1$ and ${\mathrm{m}}_2$and radii ${\mathrm{a}}_1$ and ${\mathrm{a}}_2$ respectively, kept coaxially as shown, create a zero field at P at a distance from the two centres as shown. Then

Considering a small element $d{m}_1$  in $m_1$,

${\mathrm{dE}}_1=\frac{{\mathrm{Gdm}}_1}{{\mathrm{r}}_1^2}$

Since $\sin \theta$components of $\overline{dE}$ from radially opposite elements cancel out, only $\cos \theta$ components add up. 

Total field E is ${\mathrm{E}}_1=\int {\mathrm{dE}}_1=\int \frac{{\mathrm{Gdm}}_1}{{\mathrm{r}}_1^2}\cos \theta \mathrm{d}\theta$

But $\cos \theta =\frac{\sqrt{3}}{2}$ and  ${\mathrm{E}}_1=\frac{{\mathrm{Gm}}_1}{{\mathrm{r}}_1^2}\cdot \frac{\sqrt{3}}{2}$

$(\text{where}{\mathrm{r}}_1=\sqrt{{\mathrm{a}}_1^2+{\left(\sqrt{3}{\mathrm{a}}_1\right)}^2}=2{\mathrm{a}}_1)$

$=\frac{{\mathrm{Gm}}_1}{4{\mathrm{a}}_1^2}\cdot \frac{\sqrt{3}}{2}$

$\Rightarrow$For P to have zero field

$\frac{{\mathrm{Gm}}_1\sqrt{3}}{8{\mathrm{a}}_1^2}=\frac{{\mathrm{Gm}}_2\sqrt{3}}{8{\mathrm{a}}_2^2}\Rightarrow \frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}=\frac{{\mathrm{a}}_1^2}{{\mathrm{a}}_2^2}$

Potential at P $:V_P=-(\frac{{\mathrm{Gm}}_1}{{\mathrm{r}}_1}+\frac{{\mathrm{Gm}}_2}{{\mathrm{r}}_2})$

$=-(\frac{{\mathrm{Gm}}_1}{2{\mathrm{a}}_1}+\frac{{\mathrm{Gm}}_2}{2{\mathrm{a}}_2})=\frac{-\mathrm{G}}{2}(\frac{{\mathrm{m}}_1}{{\mathrm{a}}_1}+\frac{{\mathrm{m}}_2}{{\mathrm{a}}_2})$

$PE$ is maximum at infinity, hence ($c$) is wrong.