B  

Base (bicarbonate) deficit ≈ 0.3 × body weight (kg) × (24 – measured HCO₃⁻).  
= 0.3 × 75 kg × (24 – 6) mEq/L  
= 0.3 × 75 × 18 ≈ 405 mEq, which rounds to about 400 mEq. This matches option B. [Busse, 1989, PMID 2549144]