${\mathcal{O}}(1)$.

Concerning the mixing angles, with the use of triangular matrices we have not changed the hierarchical structure of the Yukawa matrices. Hence, we can still use the usual perturbative expansion. In this way, after the transformations to left handed canonical Kähler we have, $$\begin{aligned}
\theta_{23} &=& \theta_{23}^d - \theta_{23}^u = \frac{(Y^{d}_{23})^t}{(Y^{d}_{33})^t} - \frac{(Y^{u}_{23})^t}{(Y^{u}_{33})^t} = \frac{\hat
Y^d_{23} ( 1 +{\mathcal{O}}( \varepsilon^2))}{ \hat Y^d_{33}( 1 +{\mathcal{O}}(\varepsilon^2)) } - \frac{ \hat Y^u_{23}( 1 +{\mathcal{O}}(\varepsilon^2))}{\hat Y^u_{33}( 1 +{\mathcal{\ O}}(\varepsilon^2))} \notag \\ &=& \hat \theta_{23} ( 1 +{\mathcal{O}}(\varepsilon^2))\end{aligned}$$ the discussion is identical for the $\theta_{13}$ mixing angle. The case of $\theta_{12}$ is slightly more complicated, now we have, $$\begin{aligned}
\theta_{12}^d = \frac{\displaystyle{(Y_{12}^{d})^t}} {\displaystyle{(Y_{22}^{d})^t -(Y_{23}^{d})^t (Y_{32}^{d})^t}}\end{aligned}$$ where the denominator is really the $Y_{22}^d$ element in the basis where we have already diagonalised the $2,3$ sector, and it is approximately equal to $m_s/m_b = \bar \varepsilon^2$. However, we know that both $(Y_{22}^d)^t
\leq \bar \varepsilon^2 ( 1 +{\mathcal{O}}(\varepsilon^2))$ and $(Y_{23}^d)^t (Y_{32}^d)^t \leq \bar \varepsilon^2 ( 1 +{\mathcal{O}}(\varepsilon^2))$. This means that the denominator can also be corrected only at ${\mathcal{O}}(\varepsilon^2)$, then we have, $$\begin{aligned}
\theta_{12}^d = \frac{\displaystyle{\hat Y_{12}^d (1 +{\mathcal{O}}(\varepsilon^2))}}{\displaystyle{(\hat Y_{22}^d - \hat Y_{23}^d \hat
Y_{32}^d)( 1 +{\mathcal{O}}(\varepsilon^2)) }} = \hat \theta_{12}^d ( 1 +{\mathcal{O}}(\varepsilon^2))\end{aligned}$$ doing the 