Please help me with the following question.

Dear Student,
A Seconds' pendulum is a pendulum whose time period T = 2s.
Time period of a pendulum is given by:
$T=2\pi \sqrt{\frac{l}{g}}$
So, the length of the Second's pendulum can be determined as:
$$\begin{gathered} T=2\pi \sqrt{\frac{l}{g}} \\ Squaring we get, \\ {T}^2=4{\pi }^2\frac{l}{g} \\ l=\frac{T^{2}g}{4{\pi }^2}=\frac{4g}{4{\pi }^2} \end{gathered}$$
If the acceleration due to gravity falls to 1/4th, i.e., g' = g/4
Then, the new time period would be:

$$\begin{gathered} T\text{'}=2\pi \sqrt{\frac{l}{g\text{'}}} \\ T\text{'}=2\pi \sqrt{\frac{l}{g\text{'}}}=2\pi \sqrt{\frac{2g}{4{\pi }^{2}g\text{'}}}=2\pi \sqrt{\frac{4g\times 4}{4{\pi }^{2}g}}=2\pi \sqrt{\frac{4}{{\pi }^2}}=2\times 2s=4s \end{gathered}$$

Thus,the new time period will be twice the original time period value.
Regards