= Solution
Differentiate the <displacement> components. The shear strains involving $z$ vanish because $u_z=-w_x$ and $v_z=-w_y$. The full <infinitesimal strain tensor> is
$$
\boxed{\mathcal E=z\begin{pmatrix}
-a&-b&0\\-b&-c&0\\0&0&\dfrac{\nu(a+c)}{1-\nu}
\end{pmatrix}.}
$$
In particular $\operatorname{tr}\mathcal E=(2\nu-1)(a+c)z/(1-\nu)$. Substituting the Lamé coefficients into the constitutive law gives
$$
\boxed{\tau=-\frac{Ez}{1-\nu^2}
\begin{pmatrix}a+\nu c&(1-\nu)b&0\\(1-\nu)b&c+\nu a&0\\0&0&0\end{pmatrix}.}
$$
For example $\tau_{zz}=\lambda\operatorname{tr}\mathcal E+2\mu e_{zz}=0$: the thickness-direction strain supplies exactly the plane-stress Poisson correction. Also $\tau_{xy}=-Ebz/(1+\nu)$, confirming the off-diagonal factor.
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