Based on Charle's Law, we have
[tex] \frac{V1}{V2} = \frac{T1}{T2} [/tex]
where, V and T are volume and temperature respectively.
Now, given that, V1 = 3.5 L, T1 = [tex] 20 ^{0}C[/tex] = 293 K and T2 = [tex] 100^{0}C[/tex] = 373 K
∴ V2 = [tex] \frac{(T2)X(V1)}{T1} = \frac{373 X 3.5}{293} = 4.556l [/tex]