Identify the correct use of dimensional analysis, given that $1$ in $=2\mathrm{.}540cm.$

$5.25m\frac{\text{s}\text{u}\text{v}\text{u}\text{m}\text{e}}{1(m)}\frac{.\text{v}\text{a}\text{v}\text{u}\text{m}\text{e}}{1in}=$

$5\mathrm{.}25m\frac{1(cm)}{100(m)}\frac{1in}{2\mathrm{.}540(cm)}=$

$5\mathrm{.}25m\frac{100(m)}{1(cm)}\frac{1in}{2\mathrm{.}540(cm)}=$

$5\mathrm{.}25m\frac{100(cm)}{1(m)}\frac{1in}{2\mathrm{.}540(cm)}=$

$5\mathrm{.}25m\frac{1(cm)}{100(m)}\frac{2\mathrm{.}540(cm)}{1in}=$

$5\mathrm{.}25m\frac{1(m)}{100(cm)}\frac{1in}{2\mathrm{.}540(cm)}=$

$5\mathrm{.}25m\frac{100(m)}{1(cm)}\frac{2\mathrm{.}540(cm)}{1in}=$

$5\mathrm{.}25m\frac{1(m)}{100(cm)}\frac{2\mathrm{.}540(cm)}{1in}=$