Let, V_{1} = 2 L

Given,

T_{1} = 23.4 °C or 296.4 K

T_{2} = 26.1 °C or 299.1 K

V_{2} be the volume at temperature T_{2}

If the pressure remains constant, then apply Charle’s law:

V_{1}/T_{1} = V_{2}/T_{2}

V_{2} = V_{1}T_{2}/T_{1} = 2 x 299.1 / 296.4

= 2.0182 L