Answer
4
Work Step by Step
$C_{1}=\frac{\varepsilon_{0}A}{d}$ (as separation is d and dielectric constant is 1)
$C_{2}=\frac{\kappa \varepsilon_{0}A}{2d}$ (as separation is 2d and dielectric constant is of wax is assumed to be $\kappa$)
Comparing $C_{1}$ and $C_{2}$, we find that
$C_{2}=\frac{\kappa}{2}\times C_{1}$
$\implies \kappa=\frac{2C_{2}}{C_{1}}=\frac{2\times2.6\,pF}{1.3\,pF}=4$
