#### Answer

$0$

#### Work Step by Step

Work done is given as: $W=\int_a^b F(r(t)) \dfrac{dr}{dt}(dt)$
Here, $ \dfrac{dr}{dt}=(\cos t) i-(\sin t) j +(\dfrac{1}{6}) k$
Now, $W=\int_0^{2 \pi} \cos t -\cos^2 t\sin t+2 \sin t dt$
Thus, $[\sin t+(1/3) cos^3 t-2 \cos t ]_0^{2 \pi}=0$