Answer
-66$^{\circ}$ C
Work Step by Step
$V_{1}=4250\,mL=4.25\,L$
$T_{1}=(15+273)K= 288\,K$
$P_{1}=745\, mmHg$
$V_{2}=2.50\,L$
$P_{2}=1.20\,atm= 1.20\,atm\times\frac{760\, mmHg}{1\,atm}$
$=912\, mmHg$
Inorder to get the final temperature $T_{2}$, we can use the combined gas law which is
$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$ (n is constant)
$\implies T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}=\frac{912\, mmHg\times2.50\,L\times288\,K}{745\,mmHg\times4.25\,L}$
$=207\,K=(207-273)^{\circ} C=-66^{\circ}C$
