I assume that there is no change of volume on mixing.
Then
#"Volume of 80 % + Volume of 20 % = 6 L of 70 %"#
Let #"Volume of 80 %" = x color(white)(l)"L"#; then #"Volume of 20 %" = (6 - x)color(white)(l) "L"#
Also,
#"Volume of alcohol in 80 %"color(white)(l) + "Volume of alcohol in 20 %" = "Volume of alcohol in 70 %"#
∴ #0.80 × x color(red)(cancel(color(black)("L"))) + 0.20(6 - x) color(red)(cancel(color(black)("L"))) = 0.70 × 6 color(red)(cancel(color(black)("L")))#
#0.80x + 1.2 - 0.20x = 4.2#
#0.60x = 3.0#
#x = 3.0/0.60 = 5.0#
#"Volume of 80 %" = x color(white)(l)"L" = "5 L"#
#"Volume of 20 %" = "(6 - 5) L" = "1 L"#
