Based on maximum allowable shearing stress:
$\tau_{max} = \dfrac{16TD}{\pi(D^4 - d^4)}$
$60 = \dfrac{16T(100)}{\pi (100^4 - 80^4)}$

$T = 6\,955 486.14 \, \text{N}\cdot\text{mm}$

$T = 6\,955.5 \, \text{N}\cdot\text{m}$

Based on maximum allowable angle of twist:

$\theta = \dfrac{TL}{JG}$
$0.5^\circ \left( \dfrac{\pi}{180^\circ} \right) = \dfrac{T(1000)}{\frac{1}{32}\pi (100^4 - 80^4)(83\,000)}$

$T = 4\,198\,282.97 \, \text{N}\cdot\text{mm}$

$T = 4\,198.28 \, \text{N}\cdot\text{m}$

Use the smaller torque, T = 4 198.28 N·m. *answer*

how find the value of L?? in the question I don't found the value of L?? but in solutions I see the value L-1000

Hi there Ahsan! I think they used 1000 to make the imaginary "L" into a mm unit so that the equation is suitable to the formula. You're welcome in advance!

The length is actually given if you read it carefully. It is indicated in the maximum angle of twist which is 0.5 deg/m. That is for every 1 meter (or 1000 mm), you are limited to 0.5 deg rotation.

Is there no need to convert 0.5 deg/m to deg/mm ? ^^; i tried solving it on my own but noticed you didn't so i was wondering

There is no need if your L = 1000 mm in the right side of the equation, but if your L = 1 mm, then you should convert that that angle to deg/mm.

Oh and uhh it says in the problem to <strong>Determine the maximmum torque</strong> but why do we choose the smaller one???

It actually means the maximum safe torque. If you load the larger larger torque, itwill resist the shear, but, it will twist beyond the allowable.