Find the minimum temperature at which the reaction will become spontaneous
You are only partially correct: negative Gibbs free energy at $\pu{273 K}$ is a sign of a spontaneous reaction.
However, the question asks when the reaction will become spontaneous and not to select the minimal temperature value from the multiple choices list which corresponds to the spontaneous process.

So, you were supposed to find the equilibrium temperature, e.g. when $ΔG = 0$. 
You have correctly started with the equation for the free Gibbs energy:

$$ΔG = ΔH - TΔS = 0  \quad\implies\quad T_\mathrm{eq} = \frac{ΔH}{ΔS}$$

Unknown entropy change $ΔS$ can be determined from the Gibbs free energy $ΔG_1$ at initial temperature $T_1$:

$$ΔS = \frac{ΔH - ΔG_1}{T_1}$$

Finally,

$$T_\mathrm{eq} = \frac{ΔH}{ΔS} = \frac{ΔH}{ΔH - ΔG_1}T_1 = \frac{\pu{-90 kJ mol-1}}{\pu{-90 kJ mol-1}-(\pu{-45 kJ mol-1})}\cdot\pu{273 K} = \pu{546 K}$$