how do we get these three formulas from the definition? field strength
There are two quantities B and H, both often confusingly called 'magnetic field', so when we get down to discussing the detail, most people prefer to use the terms B-field and H-field for clarity. I like to think of H as magnetic induction or driving force, and B as magnetic flux density. One  definition of B is that it's the change in the surface integral of B across the area of a linked conductor that generates a voltage in that conductor.

The relation between them is \$B=\mu_0\mu_rH\$, where \$\mu_0\$ is the permeability of free space, and \$\mu_r\$ is the relative permeability of the material with respect to a vacuum. \$\mu_r\$ has a value close to 1.000 for most ordinary materials, and up to many thousands for ferromagnetic materials.

The units of H are A/m, amps per metre. The H-field can be calculated over some path by counting the number of amps flowing through the area bounded by the path, divided by the length of the path. In the case of a straight wire, then for a circle centred on the wire with radius d, the path length is the circumference, 2*pi*d. The current is whatever is flowing through the wire, so the H-field is \$\frac{I}{2\pi d}\$. When we multiply by \$\mu_0\mu_r\$ to get to B-field, we have your first expression for B-field.

The geometry of the first situation was easy, even I can work out the length of a circumference of a circle, and the cylindrical symmetry kept it easy for relating the fields to each other.

The field in the centre of a loop is more tricky, as the effective path lengths have to be integrated over the range of paths that hug the wires out to paths that go off to infinity. The integrals are beyond me (mathematicians welcome to post a derivation in another answer), so I have to resort to that dreaded phrase 'it can be shown that' the effective path length is 0.5R, where R is the radius of the loop. The effective current is NI, where the loop has N turns, and so you get your second expression, for the B-field at the centre of the loop.

With a 'long' solenoid, the same sort of arguments and integration give your third expression.