At the transition temperature T , A leftharpoons B and Δ G^0 = 105...

At the transition temperature and $1$ atm pressure, the two states $A$ and $B$ are in equilibrium.

For a phase transition of a pure substance at $1$ atm, both states are in their standard states, so the reaction quotient $Q = 1$.

Using the relation $\Delta G = \Delta G^0 + RT \ln Q$, we get $\Delta G = \Delta G^0$.

At equilibrium, $\Delta G = 0$, which implies $\Delta G^0 = 0$.

Given $\Delta G^0 = 105 - 35\log T$, we can equate it to zero:

$105 - 35\log T = 0$

$35\log T = 105$

$\log T = 3$

$T = 10^3 = 1000$ K

Converting the temperature to Celsius:

$T(^{\circ}\text{C}) = 1000 - 273 = 727^{\circ}\text{C}$

Answer: $727$