Van der Waal's equation for n mole of gas is
[P + $\large\frac{n^2a}{V^2}$] [ V-nb] = nRT
Given
V = 4 litre
P = 11.0 atm
T = 300K
b = 0.05 litre$\;mol^{-1}$
n = 2
Thus
[ 11+ $\large\frac{2^2a}{4^2}][4-2\times0.05]$
$=2\times0.082\times300$
$a = 6.46 \;atm\;litre^2mol^{-1}$
Hence answer is (B)