Answer
$p = 2.5\times 10^{-28}~C\cdot m$
Work Step by Step
We can write a general expression for the torque:
$\tau = p \times E$
The maximum possible value of the torque is $~~\tau = p~E$
We can use the maximum value on the graph to find $p$:
$\tau_s = p~E$
$p = \frac{\tau_s}{E}$
$p = \frac{100\times 10^{-28}~N\cdot m}{40~N/C}$
$p = 2.5\times 10^{-28}~C\cdot m$
