The temperature will be 356 K.
So, you know that you must use the Clausius-Clapeyron equation. Now, you'll find this equation written is several equivalent forms, so I'll just choose one of these forms
ln(P_1/P_2) = (DeltaH_("vap"))/R * (1/T_2 - 1/T_1), where
P_1 - the vapor pressure measured at T_1;
P_2 - the vapor pressure measured at P_2;
DeltaH_("vap") - the enthalpy of vaporization;
R - the gas constant - expressed in Joules per mol K;
You have everything you need to solve for T_2. Since the pressure measured at this new temperature will be 5.00 times bigger than P_1, you can write it as P_2 = 5 * P_1 and use it in this form in the equation.
So, plug all in and you'll get
ln(P_1/(5 * P_1)) = (46340"J"/"mol")/(8.31446"J"/("mol" * "K")) * (1/T_2 - 1/"323 K")
ln(1/5) = "5573.4" * 1/T_2 - "5573.4" * 1/"323"
-1.6094 = "5573.4"/T_2 - 17.2252
15.646 = "5573.4"/T_2 => T_2 = 5573.4/15.646 = "356.2 K"
Rounded to three sig figs, the answer will be
T_2 = "356 K"
