Answer
Each jaw provides a force of 34.6N
Work Step by Step
Let F be the force in one jaw. We will sum the torques about point P to find F. Therefore;
$(+ \circlearrowleft) \sum \tau_p =0$
$11.0N*8.5cm-F*2.70cm=0$
$F=\frac{11.0N*8.5cm}{2.70cm}$
$F\approx34.6N$
Each jaw provides a force of 34.6 Newtons