If the shortest wavelength in Lyman series of hydrogen atom is $$A$$, then the longest wavelength in Paschen series of He$$^+$$ is:

For hydrogen-like atoms we use the Rydberg formula, stated first:

$$\frac{1}{\lambda}=RZ^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$$

Here $$R$$ is the Rydberg constant, $$Z$$ is the atomic number, $$n_{1}$$ is the lower (final) orbit and $$n_{2}$$ is the higher (initial) orbit.

We are told that the shortest wavelength in the Lyman series of the hydrogen atom has the value $$A$$. In the Lyman series the electron always falls to $$n_{1}=1$$. The shortest wavelength occurs when the initial level is infinitely high, that is $$n_{2}\rightarrow\infty$$. Substituting $$Z=1,\,n_{1}=1,\,n_{2}\rightarrow\infty$$ we have

$$\frac{1}{\lambda_{\text{min(Lyman, H)}}}=R\cdot1^{2}\left(\frac{1}{1^{2}}-\frac{1}{\infty^{2}}\right) =R\left(1-0\right)=R.$$

Thus

$$\lambda_{\text{min(Lyman, H)}}=\frac{1}{R}=A.$$

So we immediately obtain

$$R=\frac{1}{A}.$$

Now we turn to the Paschen series of the singly ionised helium atom $$\text{He}^{+}$$. This ion is hydrogen-like with nuclear charge $$Z=2$$. In the Paschen series the electron finally falls to $$n_{1}=3$$. The longest wavelength in any series corresponds to the smallest energy (and hence smallest frequency) transition within that series. Therefore, for Paschen, the initial level must be the next higher orbit, i.e. $$n_{2}=4$$. Using the same Rydberg formula with $$Z=2,\,n_{1}=3,\,n_{2}=4$$ we have

$$\frac{1}{\lambda_{\text{max(Paschen, He}^{+})}}=R\,(2)^{2}\left(\frac{1}{3^{2}}-\frac{1}{4^{2}}\right).$$

Now we evaluate the brackets first:

$$\frac{1}{3^{2}}-\frac{1}{4^{2}}=\frac{1}{9}-\frac{1}{16} =\frac{16-9}{144}=\frac{7}{144}.$$

Multiplying by $$Z^{2}=4$$:

$$4\left(\frac{7}{144}\right)=\frac{28}{144}=\frac{7}{36}.$$

Hence

$$\frac{1}{\lambda_{\text{max(Paschen, He}^{+})}}=R\cdot\frac{7}{36}.$$

Substituting $$R=\dfrac{1}{A}$$ obtained earlier, we get

$$\frac{1}{\lambda_{\text{max(Paschen, He}^{+})}}=\frac{1}{A}\cdot\frac{7}{36} =\frac{7}{36A}.$$

Taking the reciprocal to find the wavelength:

$$\lambda_{\text{max(Paschen, He}^{+})}=\frac{36A}{7}.$$

Hence, the correct answer is Option B.