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"#
