Two radioactive materials A and B having decay constant '7 $\lambda$ ' and ' $\lambda$ ' respectively, initially have same number of nuclei. The time taken to have the ratio of number of nuclei of material B to that of A as ' e ' is

Let $N_A(t)$ and $N_B(t)$ denote the number of nuclei of materials A and B at time $t$, with initial quantity $N_0$ for both. The decay follows $N(t) = N_0 e^{-\lambda t}$, where the decay constants are $\lambda_A = 7\lambda$ and $\lambda_B = \lambda$.

Thus, $N_A(t) = N_0 e^{-7\lambda t}$ and $N_B(t) = N_0 e^{-\lambda t}$. Given the ratio condition:

$\frac{N_B(t)}{N_A(t)} = e$

Substituting the expressions yields:

$\frac{N_0 e^{-\lambda t}}{N_0 e^{-7\lambda t}} = e^{6\lambda t} = e$

Equating exponents gives $6\lambda t = 1$, so $t = \frac{1}{6\lambda}$, corresponding to option B.

Final Answer: $\boxed{\text{B}}$