Not much can be done without knowing what [tex]\mathbf F(x,y,z)[/tex] is, but at the least we can set up the integral.
First parameterize the pieces of the contour:
[tex]C_1:\mathbf r_1(t_1)=(2\sin t_1,2\cos t_1,0)[/tex]
[tex]C_2:\mathbf r_2(t_2)=(1-t_2)(2,0,0)+t_2(3,3,3)=(2+t_2, 3t_2, 3t_2)[/tex]
where [tex]0\le t_1\le\dfrac\pi2[/tex] and [tex]0\le t_2\le1[/tex]. You have
[tex]\mathrm d\mathbf r_1=(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1[/tex]
[tex]\mathrm d\mathbf r_2=(1,3,3)\,\mathrm dt_2[/tex]
and so the work is given by the integral
[tex]\displaystyle\int_C\mathbf F(x,y,z)\cdot\mathrm d\mathbf r[/tex]
[tex]=\displaystyle\int_0^{\pi/2}\mathbf F(2\sin t_1,2\cos t_1,0)\cdot(2\cos t_1,-2\sin t_1,0)\,\mathrm dt_1[/tex]
[tex]{}\displaystyle\,\,\,\,\,\,\,\,+\int_0^1\mathbf F(2+t_2,3t_2,3t_2)\cdot(1,3,3)\,\mathrm dt_2[/tex]
