= Solution
Take a <beta-plane approximation>, $f=f_0+\beta y$ with $\beta>0$, and define the depth transport $\mathbf T=(U,V)=\int_{-H}^0\mathbf u_h\,dz$. For steady flow with no net flux through surface and bed, depth-integrated continuity gives $U_x+V_y=0$. Vertical integration of momentum gives
$$
f\hat{\mathbf k}\times\mathbf T=-\frac1{\rho_0}\nabla_h\int_{-H}^0p\,dz
+\frac{\boldsymbol\tau_w-\boldsymbol\tau_b}{\rho_0}+\nu\nabla_h^2\mathbf T.
$$
In the interior, small depth-to-horizontal aspect ratio makes vertical stress divergence the leading viscous contribution; horizontal viscous terms are neglected there. Taking the vertical component of curl eliminates pressure. Since
$$
\nabla_h\times(f\hat{\mathbf k}\times\mathbf T)=f(U_x+V_y)+\beta V=\beta V,
$$
neglecting bottom stress gives the <Sverdrup balance>,
$$
\boxed{\beta V=\frac1{\rho_0}\left(\partial_x\tau_{wy}-\partial_y\tau_{wx}\right)=\frac W{\rho_0}.}
$$
Thus the wind-stress curl sets the interior meridional depth transport, not the local surface meridional velocity. A constant-$f$ plane would have no $\beta V$ term; the use of $\beta$ in this question requires latitude-dependent $f$. For a transport <streamfunction> with $(U,V)=(-\bar\psi_y,\bar\psi_x)$, the interior relation is $\beta\bar\psi_x=W/\rho_0$.
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