Answer
$7.4\space kg$
Work Step by Step
Please see the attached image first.
Let's consider the static equilibrium of the arm.
$\sum\vec \tau_{A}\curvearrowright\space=0$
$2.7\space kg\times g\times 14\space cm+mg\times 32\space cm-T_{max}sin80^{\circ}\times3.6\space cm=0$
$32mg=760\space N\times3.6sin80^{\circ}-37.8g\space N$
$m=7.4\space kg$
Maximum mass = 7.4 kg
