a
Using the work energy theorem,

\(\left(F-m_{1} a\right) x_{1}+\left(m_{2} a\right) x_{2}-\frac{1}{2} k\left(x_{1}+x_{2}\right)^{2}=0\)

\(m_{2}\left(\frac{F}{m_{1}+m_{2}}\right)\left(x_{1}+x_{2}\right)=\frac{k}{2}\left(x_{1}+x_{2}\right)^{2}\)

\(\left(x_{1}+x_{2}\right)=m_{2}\left(\frac{F}{m_{1}+m_{2}}\right)\left(\frac{2}{k}\right)\)

\(=\frac{1(1)}{1.5}\left(\frac{2}{20}\right)\)

Further simplify the above,

\(\frac{1}{15} m =\frac{100}{15} cm\)

\(=\frac{20}{3} cm\)