The mass of the wheel is\(m = 2.8\;{\rm{kg}}\).

The radius of the wheel is\(r = 28\;{\rm{cm}}\).

The number of turns is\(\omega = 1300\;{\rm{rpm}}\).

The time is \(t = 6\;{\rm{s}}\).

The torque applied to a particular system is equivalent to the change in angular momentum in a unit of time. The final momentum in this condition will be equal to zero.

The relation of angular momentum can be written as:

\(\begin{aligned}{l}L &= I\omega \\L &= \left( {\frac{1}{2}m{r^2}} \right)\omega \end{aligned}\)

Here, Iis the moment of inertia.

On plugging the values in the above relation, you get:

\(\begin{aligned}{l}L &= \frac{1}{2}\left( {2.8\;{\rm{kg}} \times {{\left( {28\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)}^2}} \right)\left( {1300\;{\rm{rpm}} \times \frac{{2\pi \;{\rm{rad}}}}{{1\;{\rm{rev}}}} \times \frac{{1\;{\rm{min}}}}{{60\;{\rm{s}}}}} \right)\\L &= 14.9\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\end{aligned}\)

Thus, \(L = 14.9\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\) is the required angular momentum.

The relation of angular momentum can be written as:

\(\tau = \frac{{L' - L}}{t}\)

Here, \(L'\)is the final angular momentum, whose value is zero.

On plugging the values in the above relation, you get:

\(\begin{aligned}{l}\tau &= \left( {\frac{{0 - 14.9\;{\rm{kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}}}{{6\;{\rm{s}}}}} \right)\\\tau &= - 2.48\;{\rm{N}} \cdot {\rm{m}}\end{aligned}\)

Thus, \(\tau = - 2.48\;{\rm{N}} \cdot {\rm{m}}\) is the required torque.