Required Gear ratio for a DC motor
There are many contributing factors to this, most of which is related to frictional force, incline etc. If we ignore these for a moment

If we start off with \$F = ma \$ [1] 

and equally \$ T = F*r\$ [2]

Where


m = mass of vehicle
r = radius of wheels
\$T_w\$ = torque at wheels (post gearbox)
\$T_m\$ = torque at motor (pre gearbox)
F = force to accelerate the mass
\$\omega_w \$ = velocity of the wheel
\$\omega_m \$ = velocity of the motors rotor
R = gearbox ratio.


\$\frac{T_w}{r} = ma  \$ [3]

if \$T_m = -6\mu \omega_m  + 0.0059 \$  [4] 

and thus

\$T_w = -6\mu \frac{\omega_w}{R}  + 0.0059 \$ [5]

substituting back into [3]

\$\frac{-6\mu \frac{\omega_w}{R}  + 0.0059}{r} = ma \$ [6]

now it is stated the desired speed is 0.3m/s 

The circumference of the wheel is \$2 \pi r\$ &amp; thus 

\$speed  = \omega_w  r \$  [7]

thus  \$\omega_w = \frac{0.3}{ r}\$ [8]

substituting [8] into [6] and rearranging for R:

\$ R = \frac{-6\mu * 0.3}{0.00375*(0.1*1.150*0.00375-0.0059)} = 0.08777\$  or rounded up... 12:1 gearbox