Let's go through the equilibrium of gears one more time...this time, using gear diameters.

Consider the $\mathrm{FBD}$ of the small gear. Shaft $\mathrm{A}$ applies a torque to the gear in the direction shown.

The teeth of the large gear exert a force on the teeth of the small gear, opposing the direction of rotation.

Sum moments about $\mathrm{A}$ to find $\mathrm{F}$:

$$\begin{gathered} \sum _{}{\mathrm{M}}_{\mathrm{B}}=\mathrm{F}\times {\mathrm{D}}_{\mathrm{large}}/2-{\mathrm{T}}_{\mathrm{large}}=0 \\ \therefore \mathrm{F}=\frac{{\mathrm{T}}_{\mathrm{small}}}{{\mathrm{D}}_{\mathrm{small}}/2} \end{gathered}$$