Introduction:
Electropositivity is a measure of how readily an atom or ion loses electrons to form positive ions. Metals are generally electropositive because they have low ionization energies and tend to lose electrons to form positive ions. In this question, we are asked to determine the most electropositive metal among the given options.

Explanation:
To determine the most electropositive metal, we need to consider the position of each metal in the periodic table and its tendency to lose electrons.

a) Beryllium (Be):
Beryllium is located in Group 2 of the periodic table. It has an atomic number of 4 and its electronic configuration is [He] 2s². Beryllium has a relatively high ionization energy and does not readily lose its two valence electrons. Therefore, it is not as electropositive as other metals.

b) Francium (Fr):
Francium is located in Group 1 of the periodic table. It has an atomic number of 87 and its electronic configuration is [Rn] 7s¹. Francium has the lowest ionization energy among all the elements, indicating its strong tendency to lose its single valence electron and become an ion. It is the most electropositive metal among the given options.

c) Manganese (Mn):
Manganese is located in Group 7 of the periodic table. It has an atomic number of 25 and its electronic configuration is [Ar] 3d⁵ 4s². Manganese is not a typical metal in terms of electropositivity. It has a high ionization energy and tends to form positive ions less readily compared to other metals.

d) Thallium (Tl):
Thallium is located in Group 13 of the periodic table. It has an atomic number of 81 and its electronic configuration is [Xe] 4f¹⁴ 5d¹⁰ 6s² 6p¹. Thallium has a moderate ionization energy and can lose its single valence electron to form a positive ion. However, it is not as electropositive as francium.

Conclusion:
Among the given options, francium (Fr) is the most electropositive metal. It has the lowest ionization energy and readily loses its valence electron to form a positive ion.