There is no back-bonding in case of boron trihydride so it is a stronger lewis acid than boron trichloride ( even though chlorine atom is more electronegative than hydrogen atom).
Then why boron trihydride is weaker lewis acid than boron tribromide and boron tri-iodide?
Predicting the acidic strengths of $\ce{BX_3}$ using back-bonding is one plausible theory. This article1 hints us to use bond-strength data to predict the same:
The weaker Lewis acid strength of $\ce{BF3}$ compared to $\ce{BCl3}$ with respect to bases such as $\ce{NMe3}$ is explained in terms of the ligand close-packing (LCP) model. The halogen ligands remain close-packed during the formation of the complex leading to an increase in the $\pu{B−Hal}$ bond length. Because a $\ce{B−F}$ bond is stronger than a $\ce{B−Cl}$ bond, more energy is required to stretch a $\ce{B−F}$ bond than a $\ce{B−Cl}$ bond. Hence $\ce{BF3}$ is a weaker acid than $\ce{BCl3}$.
Pulling up the bond strength data, I get:
\begin{array}{c|c} \mathbf{Bond} & \mathbf{Strength (KJ/mole) } \\\hline \text{B-H} & \text{389}\\ \text{B-F} & \text{613} \\ \text{B-Cl} & \text{456}\\ \text{B-Br} & \text{377}\\ \text{B-I} & \text{?}\end{array}
Building up on these two pieces of information, it can be said that since $\ce{B-H}$ bond is stronger than $\ce{B-Br}$ and $\ce{B-I}$ (see below), $\ce{BH3}$ is a weaker acid than $\ce{BBr3}$ and $\ce{BI3}$.
I'm afraid I couldn't find the data for $\ce{B-I}$ bond, but since we can compare the bond-strengths of boron trifluorides on the basis of back-bonding theory, we can be sure that the strength of $\ce{B-I}$ bond will be $\lt \pu{377 kJ mol^-1}$.
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