A spring of force constant $k$ is cut into lengths of ratio $1:2:3$. They are connected in series and the new force constant is $k^{\prime }$. Then they are connected in parallel and force constant is $k^{\prime \prime }$. Then $k^{\prime }:k^{\prime \prime }$ is:

Spring constant is inversely proportional to length ($k\propto 1/l$).Let original length be $6L$. Cutting in $1:2:3$ gives lengths $L,2L,3L$.Individual constants: $k_1=\frac{k(6L)}{L}=6k$, $k_2=\frac{k(6L)}{2L}=3k$, $k_3=\frac{k(6L)}{3L}=2k$.Series connection: $k^{\prime }$ is the original constant (since series connection of segments of a spring is the original spring). $k^{\prime }=k$.Parallel connection: $k^{\prime \prime }=k_1+k_2+k_3=6k+3k+2k=11k$.Ratio $k^{\prime }:k^{\prime \prime }=k:11k=1:11$.