By Stokes' theorem, the integral is 0. If [tex]R[/tex] is the triangular region bounded by the given line segments composing the curve [tex]C[/tex], then
[tex]\displaystyle\int_{\partial R}\mathbf b\cdot\mathrm d\mathbf s=\iint_R\nabla\times\mathrm d\mathbf r[/tex]
where [tex]\nabla\times F=\mathrm{curl}(0,0,18)=0[/tex].
Just to verify this, we can parameterize the path by
[tex]C=C_1\cup C_2\cup C_3[/tex]
[tex]\begin{cases}C_1:=\{(4d,3dt,0)~|~0\le t\le1\}\\C_2:=\{(4d(1-t),3d(1-t),0)~|~0\le t\le1\}\\C_3:=\{(4dt,0,0)~|~0\le t\le1\}\end{cases}[/tex]
[tex]\displaystyle\int_C\mathbf b\cdot\mathrm d\mathbf s[/tex]
[tex]=\displaystyle\int_0^1(0,0,18)\cdot(0,3d,0)\,\mathrm dt+\int_0^1(0,0,18)\cdot(-4d,-3d,0)\,\mathrm dt+\int_0^1(0,0,18)\cdot(4d,0,0)\,\mathrm dt[/tex]
[tex]=0+0+0=0[/tex]
