Finding the equivalent weights of the oxide and the metal when reduced by hydrogen
Equivalents are part and parcel of Indian chemical education. I don't think college chemistry is going to be updated soon anytime there. I recently heard a complaint by a company to the chemistry chair in an American University. They said that your students do not understand normality! Please teach these concepts to your undergraduate students. Of course, normality concept sounds Greek to modern students.
Anyway let us assume we are in early 19th century and we are doing a classical experiment of chemical analysis. It is not a bad exercise, then. Balanced equations existed then, so we can write:
$$\ce{M_xO_y + yH2 -&gt; M + yH2O}$$
833 mL of hydrogen gas corresponds to 0.833/22.4 = 0.0372 moles of hydrogen. This concept existed in early 19th century.
So what is the definition of an equivalent weight of anything? It is the weight that reacts with 1 equivalent weight of hydrogen (which is half its molecular weight).
Your first step is wrong. Realize that
No. of Equivalents of $\ce{H2}$ in 833 mL are = 2 x 0.0372.
This implies that equivalents of metal oxide which are present in the reaction are also 2 x0.0372.
Equivalent weight of the metal oxide is then = 1.80 g/ 0.0744 = 24.19 or as per your textbook 24.2
Since the equivalent weight of oxygen is 8, the equivalent of metal is 24.19-8= 16.19.
Now, there is no metal whose atomic weight matches 16.19. So what if the metal is divalent = 32 (this is sulfur, not a metal). Is it possible that this is trivalent metal like Ti so  the atomic weight of that M is 16.19 x3= 48.57 which matches with titanium.
One can do back calculation to double check this using moles. I leave that for an interested reader.