Definitions of speed of light, reduced Planck's constant, and electronvolt in terms of kilograms, meters, seconds:

${\displaystyle c=299792458\ {m \over s}}$

${\displaystyle \hbar =1.054571817\times 10^{-34}\ kg{m^{2} \over s}}$

${\displaystyle eV=1.602176463\times 10^{-19}\ kg{m^{2} \over s^{2}}}$

Then solve for second:

${\displaystyle {\hbar \over eV}=6.582120268\times 10^{-16}s}$

${\displaystyle s=1.519267287\times 10^{15}{\hbar \over eV}}$

Solve for kilogram:

${\displaystyle c^{2}=8.987551787\times 10^{16}\ {m^{2} \over s^{2}}}$

${\displaystyle {eV \over c^{2}}=1.782661731\times 10^{-36}\ kg}$

${\displaystyle kg=5.609589202\times 10^{35}\ {eV \over c^{2}}}$

Solve for meter:

${\displaystyle {\hbar \over c}=3.51767294\times 10^{-43}\ kg\cdot m}$

${\displaystyle {\hbar \over c}=1.973270014\times 10^{-7}\ {eV \over c^{2}}m}$

${\displaystyle 5067730.179\ {\hbar \over c}{c^{2} \over eV}=m}$

${\displaystyle m=5067730.179\ {\hbar c \over eV}}$

Summarize results:

${\displaystyle kg=5.609589202\times 10^{35}\ {eV \over c^{2}}}$

${\displaystyle m=5067730.179\ {\hbar c \over eV}}$

${\displaystyle s=1.519267287\times 10^{15}{\hbar \over eV}}$

Dimensional conversion: ${\displaystyle kg^{x}m^{y}s^{z}\sim c^{y-2x}\hbar ^{y+z}eV^{x-y-z}}$

Now omit c and ℏ:

${\displaystyle kg=5.609589202\times 10^{35}\ eV}$

${\displaystyle m=5,067,730.179\ {1 \over eV}}$

${\displaystyle s=1.519267287\times 10^{15}{1 \over eV}}$

Dimensional conversion: ${\displaystyle kg^{x}m^{y}s^{z}\sim eV^{x-y-z}}$

Based on: http://www.physics.gla.ac.uk/~dmiller/lectures/RQM_2008.pdf, page 12