Answer
a) ${\bf 0.362}\;\rm m $
b) $\bf 0.000368\;\rm nm$
Work Step by Step
We know that the wavelengths in the hydrogen spectrum are given by
$$\lambda_{n\rightarrow m}=\dfrac{\lambda_0}{\dfrac{1}{m^2}-\dfrac{1}{n^2}}$$
where $m=1,2,3,...$, and $n=m+1,m+2,m+3,...$
and $\lambda_0=91.18\;\rm nm$
$$\lambda_{n\rightarrow m}=\dfrac{91.18\;\rm nm}{\dfrac{1}{m^2}-\dfrac{1}{n^2}}\tag 1$$
$$\color{blue}{\bf [a]}$$
Using (1); for $200\rightarrow 199$
$$\lambda_{200\rightarrow 199}=\dfrac{(91.18\times 10^{-9})}{\dfrac{1}{199^2}-\dfrac{1}{200^2}}=\color{red}{\bf 0.362}\;\rm m $$
$$\color{blue}{\bf [a]}$$
Using (1); for $199\rightarrow 2$;
$$\lambda_{199\rightarrow 2}=\dfrac{(91.18\times 10^{-9})}{\dfrac{1}{2^2}-\dfrac{1}{199^2}}=\color{blue}{\bf 3.64757\times 10^{-7}}\;\rm m $$
andfor $200\rightarrow 2$;
$$\lambda_{200\rightarrow 2}=\dfrac{(91.18\times 10^{-9})}{\dfrac{1}{2^2}-\dfrac{1}{200^2}}=\color{blue}{\bf 3.64756\times 10^{-7}}\;\rm m $$
Hence, the difference between the two wavelengths emitted;
$$\Delta \lambda=\lambda_{199\rightarrow 2}-\lambda_{200\rightarrow 2} $$
$$\Delta \lambda=\color{red}{\bf 3.67548\times 10^{-13}}\;\rm m$$
