Answer
$186KPa$
Work Step by Step
We know that
$P_2=\frac{P_1V_1T_2}{V_2T_1}$...eq(1)
Converting both $T_1$ and $T_2$ to kelvin:
$T_1=C^{\circ}+273=0+273=273K$
$T_2=C^{\circ}+273=27+273=300K$
Now we find $P_1$:
$P_1=165KPa+1.01\times10^2KPa=266KPa$
Substituting values into eq (1), we get,
$P_2=\frac{(266KPa)(1.64\times10^{-2})(300)}{(1.67\times10^{-2})(273)}=287KPa$
Thus the required gauge reading=$287KPa-1.01\times10^2KPa=186KPa$
