Consider three rods of length L _ 1 , L _ 2 and L _ 3 respectively...

Since the middle rod doesn't expand at all, the strain in the rod will be

$strain = 2\alpha ∆ T$

$stress = 2Y\left(2\alpha ∆ T\right)$

The stress is same in each rod

$4Y\alpha ∆ T = Y\dfrac{{L}_{1}\alpha ∆ T + x}{{L}_{1}} = 3Y\dfrac{{L}_{3}\left(3\alpha\right) ∆ T + x}{{L}_{3}}$

On solving for $x$, we get

$x = 3{L}_{1}\alpha ∆ T = \dfrac{5{L}_{3}\alpha ∆ T}{3}$

$⇒ \dfrac{9{L}_{1}}{{L}_{3}} = 5$