Answer:
Option C
Explanation:

F1= Net electrostatic force on anyone charge due to rest of three charges
=14πϵ0q2a2(√2+12)
F2= surface tension force =γa
If we see the equilibrium of line BC,
then 2F1cos450=F2
or √2F1=F2
or 14πϵ0q2a2(2+1√2)=γa
∴ a^{3}= \frac{1}{4 \pi \epsilon_{0}}(2+\frac{1}{\sqrt{2}})\frac{q^{2}}{\gamma}
or a= [\frac{1}{4 \pi \epsilon_{0}}(2+\frac{1}{\sqrt{2}})]^\frac{1}{3}[\frac{q^{2}}{\gamma}]^\frac{1}{3}
=k\left[\frac{q^{2}}{\gamma}\right]^{1/3}
where , k=\left\{\frac{1}{4 \pi \epsilon_{0}}(2+\frac{1}{\sqrt{2}})\right\}^{1/3}
Therefore ,N=3
Answer is 3