Explanation

(d) Given, the maximum electron density in the morning
$N_{\max} = 10^{10}\ m^{- 3}$
$f_{c}^{'} = 9\left( N_{\max} \right)^{1/2} = 9\left( 10^{10} \right)^{1/2}$
Critical frequency $f_{c}^{'} = 9 \times 10^{5}$
The maximum electron density at noon
$N_{\max} = {2 \times 10}^{10}\ m^{- 3}$
Critical frequency
$f_{c}^{''} = 9 \times \left( N_{\max}^{'} \right)^{1} = 9 \times \left( 2 \times 10^{10} \right)^{1} = 9 \times \sqrt{2} \times 10^{5}$
The ratio of critical frequency at noon and critical frequency in the morning.
$\frac{f_{c}^{''}}{f_{c}^{'}} = \frac{9 \times \sqrt{2} \times 10^{5}}{9 \times 10^{5}}$
Or $\frac{f_{c}^{''}}{f_{c}^{'}} = \sqrt{2} = 1.414$