Next, consider a $\mathrm{FBD}$ of the large gear.

Write the equilibrium equation for the sum of moments about shaft $\mathrm{B}$ to determine the torque applied to the large gear:

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