Answer
$316 \;\rm K$
Work Step by Step
Since the air has no time to exchange heat with its surroundings, the air undergoes an adiabatic process.
So,
$$P_1V_1^\gamma=P_2V_2^\gamma$$
Hence,
$$\dfrac{P_1}{P_2}=\left[ \dfrac{V_2}{V_1} \right]^\gamma$$
$$\dfrac{V_2}{V_1} =\left[ \dfrac{P_1}{P_2} \right]^\frac{1}{\gamma}\tag 1$$
Also, for an adiabatic process,
$$T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$$
Hence,
$$\dfrac{T_2}{T_1}=\left[ \dfrac{V_1}{V_2} \right]^{\gamma-1}$$
Hence,
$$ T_2=T_1\left[ \dfrac{V_1}{V_2} \right]^{\gamma-1}$$
Plugging from (1);
$$ T_2=T_1\left[ \left[\dfrac{P_1}{P_2}\right]^\frac{-1}{\gamma} \right]^{\gamma-1}=T_1\left[ \dfrac{P_1}{P_2} \right]^{\frac{1-\gamma} {\gamma}}$$
Plugging the known;
$$ T_2 =(0+273)\left[ \dfrac{60}{100} \right]^{\frac{1-1.4} {1.4}}$$
$$T_2=\color{red}{\bf 316}\;\rm K=\color{red}{\bf43}^\circ \;\rm C=\color{red}{\bf109}\;F $$
