Answer
$F=32.26N$
Work Step by Step
Area of the screw cap $A=4.10\times10^{-4}m^2$
Pressure by carbon dioxide from inside the bottle on cap
$P_{s}=1.8\times10^{5}Pa=1.8\times10^{5}N/m^2$
Suppose this creates a inside out force on cap whose magnitude is $F_{io}$
since $P=\frac{F}{A}$
so $F=PA$
so $F_{io}=1.8\times10^{5}N/m^2\times4.10\times10^{-4}m^2$
$F_{io}=7.38\times10N$
$F_{io}=73.8N$ [inside out force]...................equation(1)
Pressure of atmosphere on the bottle cap
$P_{s}=1.01325\times10^{5}Pa=1.01325\times10^{5}N/m^2$
Suppose this creates a outside to inside force on cap whose magnitude is $F_{oi}$
since $P=\frac{F}{A}$
so $F=PA$
so $F_{oi}=1.01325\times10^{5}N/m^2\times4.10\times10^{-4}m^2$
$F_{oi}=4.154325\times10N$
$F_{oi}=41.54N$ [outside ti inside force]...................equation(2)
Force calculate in equation (1)and (2) are opposite in nature. inside out force is more as compared to outside in force.
So resultant force on Cap is $F=F_{io}-F_{oi}$
$F=73.8N-41.54N=32.26N$ which is acting from inside to outside.
If screw does not support this much force cap will be open. So this much force screw thread must exert on the cap in order to keep it on the bottle.
$F=32.26N$
