Capacitance between the parallel plates of the capacitor, $C =8\, pF$ Initially, distance between the parallel plates was d and it was filled with air. Dielectric constant of air, $k =1$ Capacitance, $C$, is given by the formula,

$C=\frac{k \varepsilon_{0} A}{d}=\frac{\varepsilon_{0} A}{d} \ldots (i)$

Where, $A=$ Area of each plate

$\varepsilon_{0}=$ Permittivity of free space

If distance between the plates is reduced to half, then new distance, $d _{1}= d / 2$

Dielectric constant of the substance filled in between the plates, $k_{1}=6$

Hence, capacitance of the capacitor becomes

$C_{1}=\frac{k_{1} \varepsilon_{0} A}{d_{1}}=\frac{6 \varepsilon_{0} A}{\frac{d}{2}}=\frac{12 \varepsilon_{0} A}{d} \ldots (ii)$

Taking ratios of equations $(i)$ and $(ii)$, we obtain

$C _{1}=2 \times 6\, C =12\, C =12 \times 8 \,pF =96 \,pF$

Therefore, the capacitance between the plates is $96 \,pF$.