If Rydberg's constant is R , the longest wavelength of radiation in...

In the Paschen radiation series, the longest wavelength corresponds to the transition between $n = 3$ and $n = 4$.

The formula for the wavelength in the given transition can be written as, 

$\dfrac{1}{\lambda} = R{Z}^{2}\left(1{n}_{1}^{2} - 1{n}_{2}^{2}\right) . . . \left(1\right)$

From equation (1), it follows that

$\dfrac{1}{\lambda} = R{Z}^{2}\left(\dfrac{1}{{3}^{2}} - \dfrac{1}{{4}^{2}}\right) = R{Z}^{2}\left(\dfrac{1}{9} - \dfrac{1}{16}\right) = \dfrac{7R{Z}^{2}}{9 \times 16} ⇒ \lambda = \dfrac{144}{7R}$

 Hence, $\alpha = 144$.