Heat Transfer Coefficient Problem
Is my work correct so far?

Yes, your conversions to the units for the inputs to the equation (#1 to #3) look correct to me.

How do I simplify the units in my last step in order to then convert to a value with the desired units...?

You don't.
This sort of expression embodies a purely empirical relationship -- it has no particular basis of connection to the underlying physical relationships between these various quantities.  The reason the units of each of the input quantities are specified ($^\circ\mathrm{F}$, $\mathrm{\frac{ft}{min}}$, etc.) is because you're supposed to use just the numerical values of those quantities, after conversion to the indicated units, and substitute those into the equation. So, instead of the equation you have in #4, it would just be:
$$
H = 0.0061 \cdot \left(0.03^{0.6} + 0.038 \cdot 86\right) \cdot 246^{0.35}
$$
Then, you take the purely numerical result you obtain for $H$ from this equation and tack on the indicated units; in this case, $\mathrm{BTU \cdot {hr}^{-1} \cdot {ft}^{-2} \cdot \ {^\circ F}^{-1}}$. From there, you can carry out a straightforward conversion to the metric units asked for in the final answer.
These sorts of empirical correlations are extremely common in chemical engineering practice, where one doesn't especially care if an equation has solid theoretical underpinnings, as long as it performs reasonably well in predicting quantities of interest.
To note, while you can construct internally consistent units for the various constants in the formula, they usually don't end up having any particular useful physical interpretation. For example, since it's only possible to add two quantities together if they have the same units, the $0.6$ exponent on $X$ means that the $0.038$ coefficient on $T$ would have to have a dimensional value of $0.038\ \mathrm{{in}^{0.6} \over {^\circ F}}$, which has no particular physical significance that I can determine. I'm not even sure what sense there is to be made of units such as &quot;$\mathrm{{in}^{0.6}}$&quot;, &quot;$\left(\mathrm{ft\over min}\right)^{0.35}$&quot;, etc.