Impact of Pressure on equilibrium of Tin allotropes
First (as you already did) you need to determine the equilibrium temperature at $1\ \mathrm{bar}$. This happens to be when  $\Delta G^\circ = 0$, so that $T^\circ_{trans}= \frac{\Delta H^\circ}{\Delta S^\circ}=286.36\ \mathrm K$. 

Next you can apply the Clausius equation in integrated form 

$$ T = T^\circ_{trans} + (P-P^o) \frac{\Delta V_m^\circ}{\Delta S^\circ} $$

I obtain the following value for $\Delta V_m^\circ$:

$\Delta V_m^\circ = (1/5.766-1/7.287)\times \pu{ 118.71\times10^-6 m3/mol} = \pu{4.2973\times10^-6 m3/mol}$

Plugging this and the values you provide for $\Delta P$ ($1089\ \mathrm{bar}$) and the standard entropy changes $\Delta S_m^\circ$ gives $T=\pu{219.89 K}$. Not exactly what you say is the right answer, but, closer.