Given that,
The probability of green signal on crossing $\mathrm{X}=\mathrm{P}\left(\mathrm{G}\right)=\frac{30}{100}=\frac{3}{10}$
The probability of not a green signal on crossing $\mathrm{X}=\mathrm{P}\left({\mathrm{G}}^{\mathrm{\prime }}\right)=1-\frac{3}{10}=\frac{7}{10}$
Now the probability of finding a green signal on X on two consecutive days out of three is possible if of day 1 and 2 the signal will green and on day three it will not green or on day 1 signal is not green but on day 2 and 3 it will green,

Hence the required probability is,

$\begin{array}{l}=\mathrm{P}\left(\mathrm{G},\mathrm{G},{\mathrm{G}}^{\mathrm{\prime }}\right)+\mathrm{P}\left({\mathrm{G}}^{\mathrm{\prime }},\mathrm{G},\mathrm{G}\right)\\ =\frac{3}{10}\times \frac{3}{10}\times \frac{7}{10}+\frac{3}{10}\times \frac{3}{10}\times \frac{7}{10}\\ =\frac{63}{1000}+\frac{63}{1000}\\ =\frac{63}{500}\end{array}$
Therefore, the probability of finding a green signal on X on two consecutive days out of three is $\frac{63}{500}$