Answer
$\lambda = -1.72\times 10^{-15}~C/m$
Work Step by Step
We can find the length of the arc:
$L = \frac{40^{\circ}}{360^{\circ}}(2\pi~R)$
$L = (\frac{1}{9})(2\pi)(0.040~m)$
$L = 0.0279~m$
We can find the linear charge density:
$\lambda = \frac{q}{L}$
$\lambda = \frac{-300e}{L}$
$\lambda = \frac{(-300)(1.6\times 10^{-19}~C)}{0.0279~m}$
$\lambda = -1.72\times 10^{-15}~C/m$
