Given,
${\epsilon}_{r}=3.38\phantom{\rule{0ex}{0ex}}h=1.6$
The width of the microstrip substrate is ,
$w=\frac{1}{2{f}_{r\sqrt{{\mu}_{0}{\epsilon}_{0}}}}\sqrt{\frac{2}{{\epsilon}_{r}+1}}=\frac{{v}_{0}}{2{f}_{r}}\sqrt{\frac{2}{{\epsilon}_{r}+1}}$
Velocity of light in space,
${v}_{0}=3\times {10}^{8}m/s\phantom{\rule{0ex}{0ex}}letresonantfrequency,{f}_{r}=5Hz$
$w=\frac{{v}_{0}}{2{f}_{r}}\sqrt{\frac{2}{{\epsilon}_{r}+1}}=\frac{3\times {10}^{8}}{2\times 5\times {10}^{9}}\sqrt{\frac{2}{3.38+1}}=0.02027m\phantom{\rule{0ex}{0ex}}=2.027cm\phantom{\rule{0ex}{0ex}}ie.,\phantom{\rule{0ex}{0ex}}w=2.027cm$
Then the corresponding length is
$L\approx 0.47\frac{{\lambda}_{0}}{\sqrt{{\epsilon}_{r}}}\approx 0.47\frac{{v}_{0}}{{f}_{r}\sqrt{{\epsilon}_{r}}}\phantom{\rule{0ex}{0ex}}ie.,\phantom{\rule{0ex}{0ex}}L=0.47\frac{3\times {10}^{8}}{5\times {10}^{9}\sqrt{3.38}}=0.01533m=1.533cm$
Therefore the corresponding length is 1.533 cm