Answer
5.3
Work Step by Step
Capacitance C= 7.0$\mu F$= $7.0\times10^{-6}F$
Area A= 1.5 $m^{2}$
Plate separation d= $1.0\times10^{-5}m$
$\epsilon_{0}= 8.854\times10^{-12} F/m$
Recall that C= $\frac{K\epsilon_{0}A}{d}$.
Thus we get that the dielectric constant K= $\frac{Cd}{\epsilon_{0}A}$
$= \frac{(7.0\times10^{-6}F)(1.0\times10^{-5}m)}{(8.854\times10^{-12}F/m)(1.5m^{2})}\approx5.3$
