In first case, springs are connected in parallel, so their equivalent spring constant

$k_p=k_1+k_2$

So, frequency of this spring-block system is

$f_p=\frac{1}{2\pi } \sqrt{\frac{k_p}{m}}$

or $f_p=\frac{1}{2\pi } \sqrt{\frac{k_1+k_2}{m}}$

but $k_1=k_2=k$

$\therefore$ $f_p=\frac{1}{2\pi } \sqrt{\frac{2k}{m}}$ ..(i)

Now in second case, springs are connected in series, so their equivalent spring constant

$k=\frac{k_1{k}_2}{k_1+k_2}$

Hence, frequency of this arrangement is given by

$f_s=\frac{1}{2\pi } \sqrt{\frac{k_1{k}_2}{(k_1+k_2)m}}$

or $f_s=\frac{1}{2\pi } \sqrt{\frac{k}{2m}}$ ...(ii)

Dividing Eq. (ii) by Eq. (i), we get

$\frac{f_s}{f_p}=\frac{\frac{1}{2\pi } \sqrt{\frac{k}{2m}}}{\frac{1}{2\pi } \sqrt{\frac{2k}{m}}}=\sqrt{\frac{1}{4}}$

or $\frac{f_s}{f_p}=\frac{1}{2}$