A 340\ Å long segment of DNA molecule has 40 thymine nitrogenous...

Length of the DNA segment = $340\ \text{\AA}$

Distance between two consecutive base pairs in B-DNA = $3.4\ \text{\AA}$

Total number of base pairs = $\dfrac{340}{3.4} = 100$

Total number of nitrogenous bases = $100 \times 2 = 200$

According to Chargaff's rule, $A = T$ and $G = C$.

Given number of thymine bases ($T$) = $40$

Number of adenine bases ($A$) = $40$

Total number of $A$ and $T$ bases = $40 + 40 = 80$

Total number of $G$ and $C$ bases = $200 - 80 = 120$

Number of guanine bases ($G$) = $\dfrac{120}{2} = 60$

Answer: $60$