b
\(t = 0 \)              \(N_0 = 10^{20}\)

\(t = 1y\)             \(  N_1 = N_0e^{-\lambda(1)}\)

\(t = 2y\)             \(  N_2 =N_0e^{-\lambda(2)}\)

\(t = 3y\)            \(   N_3 = N_0e^{\lambda(3)}\)

\(\frac{{{N_2} - {N_3}}}{{{N_1} - {N_2}}} = 0.3\,\,\,\)

\( \Rightarrow  \frac{{{N_0}{e^{ - 2\lambda }} - {N_0}{e^{ - 3\lambda }}}}{{{N_0}{e^{ - \lambda }} - {N_0}{e^{ - 2\lambda }}}} = 0.3\)

\(\frac{{{N_0}{e^{ - 2\lambda }}(1 - {e^{ - \lambda }})}}{{{N_0}{e^{ - \lambda }}(1 - {e^{ - \lambda }})}} = 0.3\)

\(e^{-\lambda}  = 0.3   \)

\(\therefore N_0 - N_1  = ?\)

\(= N_0 - N_0e^{-\lambda} = N_0(1 - e^{-\lambda})\)  

\(=  10^{20} (1 - 0.3) = 0.7 \times 10^{20} = 7 \times 10^{19}\)