= Solution
Use positive <normal stress> for tension. On a small rectangular material element, $\tau_i$ is the normal force per unit area on the face perpendicular to axis $i$, and $e_i=\Delta\ell_i/\ell_i$ is the corresponding <normal strain>. <Young's modulus> $E$ is the ratio of axial <normal stress> to axial <strain> in uniaxial stress; <Poisson's ratio> $\sigma$ is minus the transverse-to-axial strain ratio in the same test. Thus isotropic <linear elasticity> gives $Ee_1=\tau_1-\sigma\tau_2-\sigma\tau_3$ and cyclic permutations. The stress symbols here follow the question; $\sigma$ denotes <Poisson's ratio>, not a stress component.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2015/paper-1-elastic-layer.png]
{title=Normal stresses on a material element and the distinct deformation fields in twist and tilt}
For $e_2=e_3=0$, the two transverse compliance equations imply $\tau_2=\tau_3=\sigma\tau_1/(1-\sigma)$. Substitution into the first gives
$$
Ee_1=\frac{(1+\sigma)(1-2\sigma)}{1-\sigma}\tau_1,\qquad
\boxed{M=\frac{E(1-\sigma)}{(1+\sigma)(1-2\sigma)}.}
$$
This <uniaxial strain modulus> is greater than the uniaxial-stress <Young's modulus> for the positive <Poisson's ratio> of interest, because the lateral <constraints> prevent free Poisson contraction.
For a small twist $\phi$, the bonded faces' relative tangential displacement at radius $r$ is $r\phi$. The layer's <shear strain> is $r\phi/h$, so the <shear stress> is $Gr\phi/h$. Integrating the <torque> gives
$$
T_V=-\frac{G\phi}h\int_A r^2\,dA
=-\frac{G\phi}h\int_0^a2\pi r^3dr
=\boxed{-\frac\pi2\frac{Ga^4}h\phi.}
$$
The minus sign denotes a restoring <torque>; the question quotes its magnitude. This uses the disk's <polar second moment of area>.
For tilt about H, choose $y$ in the plate perpendicular to H. The vertical displacement of the bonded top face is $y\theta$ to first order, giving <normal strain> $y\theta/h$. The thin layer is constrained laterally by bonding, so its leading normal traction is $My\theta/h$, not $Ey\theta/h$. Integrating its moment about H gives
$$
T_H=-\frac{M\theta}h\int_Ay^2dA
=\boxed{-\frac\pi4\frac{Ma^4}h\theta,}
$$
since the diametric <second moment of area> is $\pi a^4/4$. This is the specified thin-layer approximation to the <torsion and tilt stiffness of a bonded elastic disk>. Relaxation near the free cylindrical edge is neglected; for a nearly incompressible layer it can limit quantitative accuracy unless the thinness is sufficient.
The thin square plate's <moment of inertia> about V is $I_V=2mb^2/3$. Its two perpendicular in-plane central moments are equal and sum to $I_V$, so the diagonal H has $I_H=mb^2/3$. Dividing the two angular stiffnesses by the appropriate <moments of inertia> gives
$$
\boxed{\omega_V=\sqrt{\frac{3\pi Ga^4}{4hmb^2}},\qquad
\omega_H=\sqrt{\frac{3\pi Ma^4}{4hmb^2}}.}
$$
The geometry factors cancel in their ratio. Using the supplied relation between <shear modulus> and <Young's modulus>,
$$
R^2=\left(\frac{\omega_H}{\omega_V}\right)^2=\frac MG
=\frac{2(1-\sigma)}{1-2\sigma},\qquad
\boxed{\sigma=\frac{R^2-2}{2(R^2-1)}=0.4919\quad(R=7.9).}
$$
The result lies just below the incompressible isotropic limit $1/2$, as expected for a rubber with a large compression stiffness compared with its <shear modulus>. It is an inference within the bonded-layer model, rather than an independently measured material constant.
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