Next, consider the $\mathrm{FBD}$ for the small gear.

The small gear must be in equilibrium; therefore, the sum of moments about shaft $\mathrm{B}$ must equal zero.

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