\(\displaystyle{d}_{{{1}}}={0.64}{m}\)

\(\displaystyle{m}={1.8}{k}{g}\)

\(\displaystyle{F}_{{{1}}}={120}{N}\)

It is also given that chain sprocket:

\(\displaystyle{d}_{{{2}}}={0.09}{m}\)

\(\displaystyle\alpha={4.5}{\frac{{{r}{a}{d}}}{{{s}^{{{2}}}}}}\)

We can assume that the wheel is a hoop. Hoop moment of inertia is given by:

\(\displaystyle{I}={m}\cdot{{r}_{{{1}}}^{{{2}}}}={m}\cdot{\left({\frac{{{d}_{{{1}}}}}{{{2}}}}\right)}^{{{2}}}={1.8}\cdot{\left({\frac{{{0.64}}}{{{2}}}}\right)}^{{{2}}}={0.184}{k}{g}\ {m}^{{{2}}}\)

Torque equation:

\(\displaystyle{I}\cdot\alpha={F}_{{{2}}}\cdot{r}_{{{2}}}-{F}_{{{1}}}\cdot{r}_{{{1}}}\)

Solving it for force applied by a chain:

\(\displaystyle{F}_{{{2}}}={\frac{{{I}\cdot\alpha+{F}_{{{1}}}\cdot{r}_{{{1}}}}}{{{r}_{{{2}}}}}}={\frac{{{I}\cdot\alpha+{F}_{{{1}}}\cdot{\frac{{{d}_{{{1}}}}}{{{2}}}}}}{{{\frac{{{d}_{{{2}}}}}{{{2}}}}}}}={\frac{{{0.184}\cdot{4.5}+{120}\cdot{\frac{{{0.64}}}{{{2}}}}}}{{{\frac{{{0.09}}}{{{2}}}}}}}\)

\(\displaystyle={871.7}{N}\)